# CLEP College Algebra/Sequences and Series

In mathematics, it is important to find patterns. That is what mathematicians do almost everyday of their lives. How they determine patterns is different depending on the type of mathematics they work with. For college algebra, determining patterns is part of the curriculum. The problem below demonstrates one way we determine these patterns. Keep in mind that the problem below is an Exploration and most likely not representative of the types of problems you may see in the CLEP College Algebra exam.

Exploration 0-1: Imagine the Standard English Alphabet on a page, listed in order from left to right. After the last letter in the Standard English Alphabet, ${\displaystyle {\text{Z}}}$, define a next letter ${\displaystyle {\text{AA}}}$ that will be added to the Standard English Alphabet. Continue listing the alphabet as normal (${\displaystyle {\text{AB}},\,{\text{AC}},\,\ldots ,\,{\text{BA}},\,{\text{BB}},\ldots }$) until you get to letter ${\displaystyle {\text{ZZ}}}$, the last letter of this "New Alphabet." Assume letter ${\displaystyle {\text{A}}}$ is defined as the first letter in this "New Alphabet." At what position will you find the letter ${\displaystyle {\text{HY}}}$ in the "New Alphabet."
In the standard English Alphabet, there are ${\displaystyle 26}$ letters in total. Let us define the list of all the letters A-Z as the "Old Alphabet."

Concatenating the letter ${\displaystyle {\text{A}}}$ to ${\displaystyle {\text{A}}}$ will give us a new letter, ${\displaystyle {\text{AA}}}$. Adding this to the "Old Alphabet" obviously does not make it the old set anymore, so let us define the alphabet in which you add new letters to the old one the "New Alphabet" How many of the new letters are in this set? Well, each letter from "Old Alphabet" must get another additional letter to make the "New Alphabet" Therefore, for each letter, ${\displaystyle 26}$ letters are added to the old one (i.e. the letter "A" will have letters "A", "B", "C",...,"Z" concatenated to "A", so 26 more letters are added to each letter of the "Old Alphabet"). Since each letter is used, exactly ${\displaystyle 26^{2}{\text{ new letters}}}$ are created in the "New Alphabet" The figure below may help demonstrate this new fact.

${\displaystyle {26{\begin{cases}\overbrace {AA,\,AB,\,AC,\ldots ,\,AY,\,AZ} ^{26}\\\overbrace {BA,\,BB,\,BC,\ldots ,\,BY,\,BZ} ^{26}\\{\text{ }}\qquad \qquad \quad \;\;\vdots \\\overbrace {YA,\,YB,\,YC,\ldots ,\,YY,\,YZ} ^{26}\\\overbrace {ZA,\,ZB,\,ZC,\ldots ,\,ZY,\,ZZ} ^{26}\end{cases}}}\!}$

The left brace tells us how many letters are in the "Old Alphabet"; the top brace for each "row" tells us how many new letters are created per letter of the alphabet. Since 26 letters are created for each letter of the "Old Alphabet," add every new letter created into this subset of the "New Alphabet." Since there are ${\displaystyle 26}$ rows, and each row creates ${\displaystyle 26}$ new letters,

${\displaystyle {\underbrace {26+26+26+...+26+26} _{26}=26\cdot 26=26^{2}=676{\text{ letters in the subset of the New Alphabet.}}}}$

Remember that there are still 26 other letters in the "old alphabet" not included in the figure above. As such, add 26 to the answer above to get the total number of letters in the New Alphabet.

Given that we know the total number of letters in the "New Alphabet," we can find out what position letter HY is in this subset of the "New Alphabet." First, the letter position of H is ${\displaystyle 8}$. Second, the letter position of Y is ${\displaystyle 25}$. Since the above figure shows a table, we can find HY by looking up that "coordinate." Since the position of HY contains all the terms in that given area, we can multiply the two values to get the area (i.e. it is a rectangle that contains all those letters). This is not the final answer, however. Realize that we excluded some ${\displaystyle 7+26}$ other letters by multiplying in that block and subset; ergo, the position of the letter ${\displaystyle {\text{HY}}}$ is

${\displaystyle 8\cdot 25+(7+26)=233.}$
In the standard English Alphabet, there are ${\displaystyle 26}$ letters in total. Let us define the list of all the letters A-Z as the "Old Alphabet."

Concatenating the letter ${\displaystyle {\text{A}}}$ to ${\displaystyle {\text{A}}}$ will give us a new letter, ${\displaystyle {\text{AA}}}$. Adding this to the "Old Alphabet" obviously does not make it the old set anymore, so let us define the alphabet in which you add new letters to the old one the "New Alphabet" How many of the new letters are in this set? Well, each letter from "Old Alphabet" must get another additional letter to make the "New Alphabet" Therefore, for each letter, ${\displaystyle 26}$ letters are added to the old one (i.e. the letter "A" will have letters "A", "B", "C",...,"Z" concatenated to "A", so 26 more letters are added to each letter of the "Old Alphabet"). Since each letter is used, exactly ${\displaystyle 26^{2}{\text{ new letters}}}$ are created in the "New Alphabet" The figure below may help demonstrate this new fact.

${\displaystyle {26{\begin{cases}\overbrace {AA,\,AB,\,AC,\ldots ,\,AY,\,AZ} ^{26}\\\overbrace {BA,\,BB,\,BC,\ldots ,\,BY,\,BZ} ^{26}\\{\text{ }}\qquad \qquad \quad \;\;\vdots \\\overbrace {YA,\,YB,\,YC,\ldots ,\,YY,\,YZ} ^{26}\\\overbrace {ZA,\,ZB,\,ZC,\ldots ,\,ZY,\,ZZ} ^{26}\end{cases}}}\!}$

The left brace tells us how many letters are in the "Old Alphabet"; the top brace for each "row" tells us how many new letters are created per letter of the alphabet. Since 26 letters are created for each letter of the "Old Alphabet," add every new letter created into this subset of the "New Alphabet." Since there are ${\displaystyle 26}$ rows, and each row creates ${\displaystyle 26}$ new letters,

${\displaystyle {\underbrace {26+26+26+...+26+26} _{26}=26\cdot 26=26^{2}=676{\text{ letters in the subset of the New Alphabet.}}}}$

Remember that there are still 26 other letters in the "old alphabet" not included in the figure above. As such, add 26 to the answer above to get the total number of letters in the New Alphabet.

Given that we know the total number of letters in the "New Alphabet," we can find out what position letter HY is in this subset of the "New Alphabet." First, the letter position of H is ${\displaystyle 8}$. Second, the letter position of Y is ${\displaystyle 25}$. Since the above figure shows a table, we can find HY by looking up that "coordinate." Since the position of HY contains all the terms in that given area, we can multiply the two values to get the area (i.e. it is a rectangle that contains all those letters). This is not the final answer, however. Realize that we excluded some ${\displaystyle 7+26}$ other letters by multiplying in that block and subset; ergo, the position of the letter ${\displaystyle {\text{HY}}}$ is

${\displaystyle 8\cdot 25+(7+26)=233.}$

What were we doing in the problem above? Essentially, we were simply trying to find what position some "term" is in. Does it not intrigue you to see math try to find a position of some "term" in a list? We have problems like these as mathematicians because the patterns underlying a "list" of numbers can help us determine new facts of mathematics. After all, what were we doing when using functions? We were trying to find a number using a pattern (the function). Unlike the previous sections, however, we were not given a formula. Luckily enough, it is not difficult to make a formula for a given "list" of numbers. Before we dive into these new problems. It helps to establish definitions.

Definition.

A sequence is a list of elements, such as numbers, figures, or letters, that is generally written in some pattern.

Objects in Sequences.

A term is an item found within a sequence.

There is a little disclaimer to get out of the way before we try to solve some problems. First, a sequence can have no pattern. However, for our purposes, we will not count any list of numbers in which no pattern exists. Second, even in simple sequences, numbers of any kind can be named if they follow after a sequence rule. For example, here's a sequence in which the rule is to list prime numbers: ${\displaystyle {\left\{{\mathit {2}},{\mathit {3}},{\mathit {5}},{\mathit {7}},{\mathit {11}},\ldots \right\}}}$ If you were like most people, you would probably name the prime numbers in order. However, you could perhaps finish the sequence like this: ${\displaystyle {\left\{{\mathit {2}},{\mathit {3}},{\mathit {5}},{\mathit {7}},{\mathit {11}},{\mathit {13}},{\mathit {23}},{\mathit {2011}},{\mathit {(2^{82,589,933}-1)}},\ldots \right\}}}$ For our purpose of standardization, we will follow a pattern by stating what number you must find first or identify the pattern.

Let's begin exploring the world that is Sequences and Series.

## Sequences

As you already know, a sequence is a list of objects that generally follow a pattern. However, the type of pattern that is described will classify sequences into either arithmetic sequences or geometric sequences. Each will be explored in depth within the next sections.

### Arithmetic Sequences

Location of a term.

An index is the location of a term within a sequence, usually denoted by ${\displaystyle i}$ or ${\displaystyle k}$.

Definition of Arithmetic Sequence.

An arithmetic sequence is a sequence ${\displaystyle {\left\{x_{1},x_{2},\ldots ,x_{k},x_{k+1},\ldots \right\}}}$ in which an added real number ${\displaystyle d}$, called the difference, is added to each successive term, except the first term, ${\displaystyle x_{1}}$, such that the sequence forms ${\displaystyle {\left\{x_{1},x_{1}+d,x_{1}+2d,x_{1}+3d,\ldots \right\}}}$ in a one-to-one correspondence.

An example could perhaps help you figure with the formal definition above: The sequence ${\displaystyle {\left\{2,4,6,8,10,\ldots \right\}}}$ has a one-to-one correspondence with the general sequence ${\displaystyle {\left\{x_{1},x_{2},x_{3},x_{4},x_{5},\ldots \right\}}}$ since the first term ${\displaystyle x_{1}=2}$, second term ${\displaystyle x_{2}=4}$, third term ${\displaystyle x_{3}=6}$, and so on. The difference is the amount added to each previous term to get the new term. For ${\displaystyle x_{2}}$, ${\displaystyle x_{1}+d=x_{2}}$ or ${\displaystyle 2+d=4}$. Solving for ${\displaystyle d}$ is the difference of the two terms. In this example, the difference is ${\displaystyle d=2}$. This is how we define an arithmetic sequence.

#### Recursive Formula

Often times, we want to generate a sequence using a formula (we are mathematicians, after all, and we like to study sequences to see if there are any general patterns). If we want to find ${\displaystyle x_{i}}$, we may use the following formula:

${\displaystyle x_{i}=x_{i-1}+d.}$

However, the above formula could describe any sequence that has that general pattern. To fix this, we need to describe the first term as well when using the formula above. There are two ways to describe this formula:

1. We have it horizontally deliniated: ${\displaystyle x_{1}=a;x_{i}=x_{i-1}+d}$.
2. We have it vertically deliniated: ${\displaystyle {\begin{cases}x_{1}=a\\x_{i}=x_{i-1}+d\end{cases}}}$

To save space, we will horizontally deliniate formulas for arithmetic sequences in this WikiBook.

Formulas in which the first term is identified along with an equation in which the previous term is added by ${\displaystyle d}$ to get the next term is called a recursive formula.

A recursive formula is a formula that describes how to get a term or many terms of a sequence by stating the starting value or values and each previous term or terms and adding difference ${\displaystyle d}$ to each subsequent term.

Recursive Formula for an Arithmetic Sequence

Given an initial value ${\displaystyle a}$ and constant difference ${\displaystyle d}$, the ${\displaystyle i}$th term in an arithmetic sequence is given by the previous ${\displaystyle (i-1)}$th term of the sequence:

${\displaystyle x_{1}=a;x_{i}=x_{i-1}+d}$ OR
${\displaystyle {\begin{cases}x_{1}=a\\x_{i}=x_{i-1}+d\end{cases}}}$
 Example 1.1.1.a: Find the ${\displaystyle 7}$th term to the arithmetic sequence ${\displaystyle {\left\{19,18.25,17.5,16.75,16,\ldots \right\}}}$ To determine the next term, we first need to find the difference ${\displaystyle d}$. Note that an arithmetic sequence will have the next term add ${\displaystyle d}$ to the previous term. Since that is how the arithmetic sequence works, ${\displaystyle 19+d=18.25}$ is a valid way to find the difference between two terms. Solving for ${\displaystyle d}$, we find the constant difference is ${\displaystyle d=-{3 \over 4}=-0.75}$. Since the difference is the same for each given term in the sequence, we can find the 7th term by adding ${\displaystyle d=-0.75}$ to ${\displaystyle 16}$, which gives us ${\displaystyle 16-0.75=15.25}$. Finally, add ${\displaystyle d}$ to the next term to get the final answer: ${\displaystyle 15.25-0.75=14.5.}$

There are many reasons why it is more important to have a recursive formula. It is not always slow; it may be easier to understand. The next example shows why this is exactly true

 Example 1.1.1.b: Write the recursive formula to the sequence ${\displaystyle {\left\{1,1,2,3,5,7,13,\ldots \right\}}}$. Many of you will perhaps know this famous pattern as the Fibonacci Sequence. For those of you who do not know this sequence, the way we determine the next term is by using the previous terms and adding them together. In our notation, we would say that the term ${\displaystyle a}$ at index ${\displaystyle i}$, ${\displaystyle a_{i}}$ is equivalent to ${\displaystyle a_{i-2}+a_{i-1}}$. Remember, however, we are not done. If a mathematician saw the sequence ${\displaystyle {\left\{2,4,6,10,16,26,42,\ldots \right\}}}$, he (or she) would determine that ${\displaystyle a_{i}=a_{i-2}+a_{i-1}}$ also describes that sequence. Therefore, we must list the first two terms because listing only the first term would not allow us to get the next term. This means our final answer is ${\displaystyle a_{1}=1,a_{2}=1;a_{i}=a_{i-2}+a_{i-1}.}$

Note that Example 1.1.b is not an example of an arithmetic sequence. Your next exploration will be to determine why this is true. Along with that, you will use your critical thinking skills to argue for or against something in the explorations after that one.

Exploration 1-1: Explain why the Fibonacci Sequence is not considered arithmetic. Determine how to change the formula for the Fibonacci Sequence so that it becomes an arithmetic recursive formula; explain why must the formula be written that way.
A term ${\displaystyle x}$ at index ${\displaystyle i}$ is not added by a common difference ${\displaystyle d}$, where ${\displaystyle d}$ must remain constant. By definition, the Fibonacci Sequence does not have a constant difference to get to the next term because the formula ${\displaystyle a_{1}=1,a_{2}=1;a_{i}=a_{i-2}+a_{i-1}.}$ does not have a constant difference. Ergo, the Fibonacci Sequence formula is not arithmetic, but it is recursive. To make it arithmetic, identify only the first term because we can find the second term by using the constant difference ${\displaystyle d}$, then add the previous term of ${\displaystyle i}$, ${\displaystyle a_{i-1}}$, by constant difference ${\displaystyle d}$: ${\displaystyle a_{1}=1;a_{i}=a_{i-1}+d}$.
A term ${\displaystyle x}$ at index ${\displaystyle i}$ is not added by a common difference ${\displaystyle d}$, where ${\displaystyle d}$ must remain constant. By definition, the Fibonacci Sequence does not have a constant difference to get to the next term because the formula ${\displaystyle a_{1}=1,a_{2}=1;a_{i}=a_{i-2}+a_{i-1}.}$ does not have a constant difference. Ergo, the Fibonacci Sequence formula is not arithmetic, but it is recursive. To make it arithmetic, identify only the first term because we can find the second term by using the constant difference ${\displaystyle d}$, then add the previous term of ${\displaystyle i}$, ${\displaystyle a_{i-1}}$, by constant difference ${\displaystyle d}$: ${\displaystyle a_{1}=1;a_{i}=a_{i-1}+d}$.
Exploration 1-2: Let the first two terms of a sequence be ${\displaystyle x_{1}=a}$ and ${\displaystyle x_{2}=b}$ where ${\displaystyle a\neq b}$.
• Argue either for OR against the claim that the recursive formula ${\displaystyle x_{1}=a,x_{2}=b;x_{i}=x_{i-2}+d}$ is arithmetic and recursive.
• If you disagree that ${\displaystyle x_{1}=a,x_{2}=b;x_{i}=x_{i-2}+d}$ is arithmetic and recursive, write a formula that is arithmetic and recursive.
• If you agree that ${\displaystyle x_{1}=a,x_{2}=b;x_{i}=x_{i-2}+d}$ is arithmetic and recursive, explain why it is so.

Finally, if ${\displaystyle x_{1}=a}$, ${\displaystyle x_{2}=b}$, and ${\displaystyle a=b}$,

• argue either for OR against the idea that the formula ${\displaystyle x_{1}=a,x_{2}=b;x_{i}=x_{i-2}+d}$ is arithmetic and recursive.
* The sequence generated by using the formula is ${\displaystyle {\left\{a,b,\left(a+d\right),\left(b+d\right),\left(a+2d\right),\left(b+2d\right),\left(a+3d\right),\left(b+3d\right),\ldots \right\}}}$ For a sequence to be arithmetic, the common difference ${\displaystyle d}$ must be constant. While the common difference is constant, the sequence must be in the form ${\displaystyle {\left\{a,\left(a+d\right),\left(a+2d\right),\left(a+3d\right),\left(a+4d\right)\ldots \right\}}}$; otherwise, it is not arithmetic. Therefore, the recursive formula ${\displaystyle x_{1}=a,x_{2}=b;x_{i}=x_{i-2}+d}$ is not arithmetic. To make it an arithmetic recursive formula, change the previous formula such that ${\displaystyle x_{1}=a;x_{i}=x_{i-1}+d}$.
• If ${\displaystyle a=b}$, then the sequence generated by using the formula ${\displaystyle x_{1}=a,x_{2}=b;x_{i}=x_{i-2}+d}$ is ${\displaystyle {\left\{b,b,\left(b+d\right),\left(b+2d\right),\left(b+3d\right),\left(b+4d\right),\left(b+5d\right),\ldots \right\}}}$ However, for a sequence to be arithmetic, each term must be added by the constant difference. Since the term in index ${\displaystyle i=2}$ is not added by ${\displaystyle d}$, the formula is not arithmetic.
* The sequence generated by using the formula is ${\displaystyle {\left\{a,b,\left(a+d\right),\left(b+d\right),\left(a+2d\right),\left(b+2d\right),\left(a+3d\right),\left(b+3d\right),\ldots \right\}}}$ For a sequence to be arithmetic, the common difference ${\displaystyle d}$ must be constant. While the common difference is constant, the sequence must be in the form ${\displaystyle {\left\{a,\left(a+d\right),\left(a+2d\right),\left(a+3d\right),\left(a+4d\right)\ldots \right\}}}$; otherwise, it is not arithmetic. Therefore, the recursive formula ${\displaystyle x_{1}=a,x_{2}=b;x_{i}=x_{i-2}+d}$ is not arithmetic. To make it an arithmetic recursive formula, change the previous formula such that ${\displaystyle x_{1}=a;x_{i}=x_{i-1}+d}$.
• If ${\displaystyle a=b}$, then the sequence generated by using the formula ${\displaystyle x_{1}=a,x_{2}=b;x_{i}=x_{i-2}+d}$ is ${\displaystyle {\left\{b,b,\left(b+d\right),\left(b+2d\right),\left(b+3d\right),\left(b+4d\right),\left(b+5d\right),\ldots \right\}}}$ However, for a sequence to be arithmetic, each term must be added by the constant difference. Since the term in index ${\displaystyle i=2}$ is not added by ${\displaystyle d}$, the formula is not arithmetic.
• Exploration 1-3: Let the first term ${\displaystyle x_{1}=a}$. Argue either for OR against the idea that sequence ${\displaystyle {\left\{x_{1},x_{2},x_{3},\ldots \right\}}}$ is arithmetic if you change it to ${\displaystyle {\left\{a,a,a,\ldots \right\}}}$
Since ${\displaystyle {x_{1}=a;x_{i}=x_{i-1}+d}}$ is the arithmetic recursive formula, ${\displaystyle a=a+d}$ is valid. The only way for ${\displaystyle a=a+d}$ to be true is by making ${\displaystyle d=0}$. However, since you are adding each term by using a constant difference ${\displaystyle d}$, and you are using the previous term to do so, the new sequence forms: ${\displaystyle {a,a+1\cdot 0,a+2\cdot 0,a+3\cdot 0,a+4\cdot 0,\ldots }}$ By definition, the sequence formed is arithmetic.
Since ${\displaystyle {x_{1}=a;x_{i}=x_{i-1}+d}}$ is the arithmetic recursive formula, ${\displaystyle a=a+d}$ is valid. The only way for ${\displaystyle a=a+d}$ to be true is by making ${\displaystyle d=0}$. However, since you are adding each term by using a constant difference ${\displaystyle d}$, and you are using the previous term to do so, the new sequence forms: ${\displaystyle {a,a+1\cdot 0,a+2\cdot 0,a+3\cdot 0,a+4\cdot 0,\ldots }}$ By definition, the sequence formed is arithmetic.

#### Direct Formula

By now, you may be wondering if there is a way to find the term of an arithmetic sequence directly. Well, there is. Before giving you the formula, let us go through the motions for our general recursive arithmetic formula ${\displaystyle {x_{1}=a;x_{k}=x_{k-1}+d}}$. Let's chart the recursive arithmetic formula.

${\displaystyle {\begin{array}{|c|c|}k&x\\\hline 1&a\\\hline 2&a+d\\\hline 3&a+2d\\\hline 4&a+3d\\\hline \vdots &\vdots \end{array}}\!}$

If you think about it, the table above is basically a linear function, although starting at ${\displaystyle k=1}$ instead of ${\displaystyle k=0}$. Write out the function as ${\displaystyle x=a+kd}$. We get near our answer. Our independent variable ${\displaystyle k}$ is horizontally translated by ${\displaystyle 1}$ to the right, so ${\displaystyle x=a+(k-1)d}$ is our function. In fact, we found our direct relationship. Rewrite it the way we normally write it and we found our direct formula: ${\displaystyle x_{k}=a+(k-1)d}$.

A direct formula describes how to get a term or many terms of a sequence by only using a formula that directly finds a particular value, without stating the first term or terms of a sequence.

Direct Formula for an Arithmetic Sequence

Given an initial value ${\displaystyle a}$ and constant difference ${\displaystyle d}$, the ${\displaystyle i}$th term in an arithmetic sequence is given by the direct formula:

${\displaystyle x_{i}=a+(i-1)d.}$

 Example 1.1.2.a: An arithmetic sequence ${\displaystyle {\left\{22.4,29.2,36,42.8,49.6,\ldots \right\}}}$ is discovered. What is the ${\displaystyle 450}$th term in the sequence? As always, before we can determine the answer, we need to find the "rate of change" of our sequence. Since ${\displaystyle a=22.4}$, we know that ${\displaystyle x_{i}=22.4+(i-1)d}$. The second term is ${\displaystyle 29.2}$, so when ${\displaystyle i=2}$, ${\displaystyle 29.2=22.4+(2-1)d}$. Of course, now we can solve for ${\displaystyle d}$: ${\displaystyle 29.2=22.4+(2-1)d}$; ${\displaystyle 29.2=d}$; ${\displaystyle 6.8=d}$. Since we now know the common difference, we can find the smallest ${\displaystyle 450}$th term in the sequence. By using our direct formula for an arithmetic sequence, we can find the index in which it is possible. Since ${\displaystyle x_{i}=22.4-6.8(i-1)}$, we find the ${\displaystyle 450}$th term by substituting ${\displaystyle i=450}$. Ergo, ${\displaystyle x_{450}=22.4-6.8(450-1)}$. Solve for ${\displaystyle x_{450}}$ to get the final answer. ${\displaystyle x_{450}=22.4-6.8(450-1)}$; ${\displaystyle x_{450}=22.4-6.8(449)}$; ${\displaystyle x_{450}=22.4-3121.2}$; ${\displaystyle x_{450}=-3030.8.}$

The example above would be a routine, straightforward problem in the CLEP College Algebra exam. However, as practice makes perfect, we will also have non-routine problems that involve thorough understanding of the topic and concepts and skills learned, which will make up 50% of the exam. This is why it is important to do the explorations. While they may not be on a CLEP exam, they are vital in making you think like a mathematician. The next problem will be non-routine problem.

 Example 1.1.2.b: What is the smallest index needed to find a negative term in the arithmetic sequence ${\displaystyle {\left\{22.4,21.2,20,17.6,16.4,\ldots \right\}}}$? As always, before we can determine the answer, we need to find the "rate of change" of our sequence. Since ${\displaystyle a=22.4}$, we know that ${\displaystyle x_{i}=22.4+(i-1)d}$. We know that the second term is ${\displaystyle 21.2}$, so when ${\displaystyle i=2}$, ${\displaystyle 21.2=22.4+(2-1)d}$. Of course, now we can solve for ${\displaystyle d}$: ${\displaystyle 21.2=22.4+(2-1)d}$; ${\displaystyle 21.2=22.4+d}$; ${\displaystyle -1.2=d}$. Since we now know the common difference, we can find the smallest term needed to reach a negative number in our sequence. By using our direct formula for an arithmetic sequence, we can find the index by solving for ${\displaystyle i}$. Since ${\displaystyle x_{i}=22.4-1.2(i-1)}$, we find the minimal index in which it is possible to have a number less than zero. Ergo, ${\displaystyle 0\geq 22.4-1.2(i-1)}$. Now all we have to do is solve for ${\displaystyle i}$. ${\displaystyle 22.4-1.2(i-1)\leq 0}$; ${\displaystyle -1.2(i-1)\leq -22.4}$; ${\displaystyle i-1\geq {-22.4 \over -1.2}}$; ${\displaystyle i\geq {22.4 \over 1.2}+1}$; ${\displaystyle i\geq 19.{\overline {6}}}$. Since index ${\displaystyle i}$ must be greater than ${\displaystyle 19.{\overline {6}}}$, the minimal index required to find the term that is negative is at ${\displaystyle i=20.}$

#### CLEP Practice Problems: Check Your Understanding

1 In the arithmetic series ${\displaystyle {\left\{2,\,6,\,10,\,14,\ldots ,\,398\right\}}}$ in which ${\displaystyle 398}$ is the last term, how many terms are in the sequence?

 terms. (Note: typing a comma will give you a decimal, so do not type a comma.)

2 A school district has set-up a system whereby each school week day (days Monday through Friday) is categorized into either an A-day or B-day. Abigail is tired of having her pencils not returned to her. She decided to devise a system in which each student will record their first and last name. The pencils are free for anyone to borrow. However, should a student forget to return a pencil to her before the end of class, a student is charged ${\displaystyle 25}$ cents first. After, another ${\displaystyle 25}$ cents are charged per cycled block day the pencil is not returned – e.g. if a student accidentally forgot to give back Abigail's pencil, they will be fined the next A-day after the B-day. Joseph had forgotten to return the pencil for ${\displaystyle 60}$ school week days. How much does Joseph owe Abigail, according to her system?

 $${\displaystyle 15.00}$$${\displaystyle 14.75}$ $${\displaystyle 7.75}$$${\displaystyle 7.50}$ \$${\displaystyle 7.25}$

3 The following arithmetic sequence is listed: ${\displaystyle {\left\{-{1 \over 3},\,0,\,{1 \over 3},\,{2 \over 3},\,1,\,1{1 \over 3},1{2 \over 3},\,2,\ldots \right\}}}$ Which of the following is the formula that represents the sequence directly using any index ${\displaystyle i}$ to find term ${\displaystyle x_{i}}$?

 ${\displaystyle x_{i}={1 \over 3}-{1 \over 3}(i+1)}$ ${\displaystyle x_{i}={1 \over 3}-{1 \over 3}(i-1)}$ ${\displaystyle x_{i}=-{1 \over 3}+{1 \over 3}i}$ ${\displaystyle x_{i}=-{1 \over 3}+{1 \over 3}(i-1)}$ ${\displaystyle x_{i}=-{1 \over 3}+{1 \over 3}(i+1)}$

4 If the first term is ${\displaystyle x}$ in a sequence, the last ${\displaystyle 501}$st term of the sequence is ${\displaystyle 9,000}$, and the constant difference is ${\displaystyle 20}$, what is the second term of the sequence?

 ${\displaystyle -1,000}$ ${\displaystyle -980}$ ${\displaystyle -1,020}$ ${\displaystyle 1,000}$ ${\displaystyle 980}$

5 Given ${\displaystyle a_{i}}$ is the term of a sequence at index ${\displaystyle i}$, the first term of a sequence is ${\displaystyle x}$, and the constant difference is ${\displaystyle d}$, which of the following must be TRUE? Indicate all such statements.

 ${\displaystyle a_{1}=x;a_{i}=x+d(i-2)+d}$ ${\displaystyle {x-a_{i} \over d}-1=i}$ ${\displaystyle a_{i}-x+d=di}$

### Geometric Sequence

Definition of Geometric Sequence.

A geometric sequence is a sequence ${\displaystyle {\left\{x_{1},x_{2},\ldots ,x_{k},x_{k+1},\ldots \right\}}}$ in which a multiplied real number ${\displaystyle r}$, called the common ratio, is multiplied to each successive term, except the first term, ${\displaystyle x_{1}}$, such that the sequence forms ${\displaystyle {\left\{x_{1},x_{1}r,x_{1}r^{2},x_{1}r^{3},\ldots \right\}}}$

As always, if you are unable to understand, try a few examples of numbers to think of in your head. Let the common ratio ${\displaystyle r=2}$ and let ${\displaystyle x_{1}=1.5}$. The next term ${\displaystyle {x_{2}=x_{1}r}}$, so ${\displaystyle x_{2}=1.5\cdot 2=3}$. If you keep the pattern going for each term of the sequence, you would get the following: ${\displaystyle {\left\{1.5,3,6,12,24,48,\ldots \right\}}}$

#### Recursive Formula

As with the arithmetic formula, you can find the recursive formula and the direct formula for a geometric sequence. Since every term is multiplied by common ratio ${\displaystyle r}$, let any term of index ${\displaystyle i}$ be represented by ${\displaystyle x_{i}}$. To find the next term requires knowing the previous term. Ergo, ${\displaystyle x_{i}=rx_{i-1}}$

Recursive Formula for a Geometric Sequence

Given an initial value ${\displaystyle a}$ and constant difference ${\displaystyle d}$, the ${\displaystyle i}$th term in an arithmetic sequence is given by the previous ${\displaystyle (i-1)}$th term of the sequence:

${\displaystyle x_{1}=a;x_{i}=rx_{i-1}}$ OR
${\displaystyle {\begin{cases}x_{1}=a\\x_{i}=rx_{i-1}\end{cases}}}$

#### Direct Formula

As with the arithmetic sequence, Let us chart the recursive geometric formula. ${\displaystyle {x_{1}=a;x_{k}=rx_{k-1}}}$.

${\displaystyle {\begin{array}{|c|c|}k&x\\\hline 1&a\\\hline 2&ar\\\hline 3&ar^{2}\\\hline 4&ar^{3}\\\hline \vdots &\vdots \end{array}}\!}$

If you think about it, the table above is basically an exponential function, ${\displaystyle f(k)=ab^{k}}$, starting at ${\displaystyle k=1}$. Write out the function as ${\displaystyle x=ar^{k}}$. We get near our answer. Our independent variable ${\displaystyle k}$ is translated to the right ${\displaystyle 1}$ unit, so ${\displaystyle x=ar^{k-1}}$ is our function. In fact, we found our direct relationship. Rewrite it the way we normally write it and we found our direct formula: ${\displaystyle x_{k}=ar^{k-1}}$.

Direct Formula for a Geometric Sequence

Given an initial value ${\displaystyle a}$ and common ratio ${\displaystyle r}$, the ${\displaystyle i}$th term in an geometric sequence is given by the direct formula:

${\displaystyle x_{i}=ar^{i-1}}$

 Example 1.2.2.a: A geometric sequence ${\displaystyle {\left\{22.4,11.2,5.6,2.8,1.4,\ldots \right\}}}$ is discovered. What is the ${\displaystyle 18}$th term in the sequence? To find the answer, we need to know the common ratio ${\displaystyle r}$ of the sequence above. Pick any arbitrary term in the sequence and apply it to the direct geometric formula: ${\displaystyle x_{2}=ar^{2-1}}$. Find ${\displaystyle r}$: ${\displaystyle 11.2=22.4r.}$ ${\displaystyle r=0.5={1 \over 2}.}$ Knowing the value of ${\displaystyle r}$, you can find the ${\displaystyle 18}$th term in the sequence by using the direct formula: ${\displaystyle x_{18}=22.4\left({\frac {1}{2}}\right)^{18-1}}$ ${\displaystyle x_{18}=22.4\left(0.00000762939\right)}$ ${\displaystyle x_{18}=0.00017089843}$

As always, these examples are things you can work through yourself or follow along so that you can see how to do a problem.

Exploration 1-4: Let the first term ${\displaystyle x_{1}=a}$. Argue either for OR against the idea that sequence ${\displaystyle {a,-2a,4a,\ldots }}$ is geometric.
A geometric sequence must have the terms be multiplied by a common ratio ${\displaystyle r}$. If the sequence were arithmetic, a term must be added such that adding the previous term by a constant difference ${\displaystyle d}$ will result in the alternation of negative and positive terms. Since that is impossible, you must multiply the terms in the sequence above by a common ratio ${\displaystyle r}$.
A geometric sequence must have the terms be multiplied by a common ratio ${\displaystyle r}$. If the sequence were arithmetic, a term must be added such that adding the previous term by a constant difference ${\displaystyle d}$ will result in the alternation of negative and positive terms. Since that is impossible, you must multiply the terms in the sequence above by a common ratio ${\displaystyle r}$.
Exploration 1-5: A power function is a function ${\displaystyle f}$ that has the power term ${\displaystyle p}$ be any real number ${\displaystyle \mathbb {R} }$ or ${\displaystyle p\in \mathbb {R} }$, where ${\displaystyle f(x)=ax^{p}}$. Let ${\displaystyle p={1 \over 2}}$ and ${\displaystyle a=2}$. If ${\displaystyle f(x)=ax^{p}}$,
• multiply each value of ${\displaystyle x}$ by a common ratio ${\displaystyle r}$. Determine whether the range will also be a geometric sequence through example using any value of ${\displaystyle r}$.
• prove that multiplying the domain, ${\displaystyle x}$, by a common ratio ${\displaystyle r}$ will give a range that is geometric (where the common difference for the domain is ${\displaystyle r_{f})}$.
* Since the domain of ${\displaystyle f}$ must be a geometric sequence multiplied by a common ratio ${\displaystyle r}$, multiply each value of ${\displaystyle x}$ by ${\displaystyle r=2}$. The following geometric sequence is formed: since ${\displaystyle x\cdot 2}$, given ${\displaystyle f(2x)=2{\sqrt {2x}}}$, ${\displaystyle f(2x)}$ will form ${\displaystyle {2{\sqrt {2}},\,4,\,2{\sqrt {6}},\,2{\sqrt {8}},\ldots }}$. Each sequence must have values multiplied by ${\displaystyle {\sqrt {2}}}$. As such, the range of ${\displaystyle f}$ must be geometric for this given ${\displaystyle r}$ value.
• The question is asking whether or not ${\displaystyle f(xr)}$ will give a range for ${\displaystyle f(x)}$ such that you multiply by a constant. Let's define that constant as ${\displaystyle r_{f}}$. Let ${\displaystyle f(x)=f\left(x_{1}\right)}$. Since ${\displaystyle f\left(x_{1}\right)=a\left(x_{1}\right)^{p}}$, if ${\displaystyle x_{2}=x_{1}r}$, then ${\displaystyle f\left(x_{2}\right)=a\left(x_{2}\right)^{p}}$. Since ${\displaystyle x_{2}=x_{1}r}$, function ${\displaystyle f\left(x_{2}\right)=a\left(x_{1}r\right)^{p}}$. Since any term ${\displaystyle (ab)^{m}}$ will give ${\displaystyle a^{m}b^{m}}$, function ${\displaystyle f\left(x_{2}\right)=ax_{1}^{p}r^{p}}$. Since ${\displaystyle ax_{1}^{p}=f\left(x_{1}\right)}$, function ${\displaystyle f\left(x_{2}\right)=f\left(x_{1}\right)r^{p}}$. Since ${\displaystyle r}$ and ${\displaystyle p}$ are constant, the expression ${\displaystyle r^{p}}$ will be constant. Since you are multiplying the range, ${\displaystyle f\left(x_{1}\right)}$, by a common ratio (otherwise known as a geometric sequence), given the constant ${\displaystyle r^{p}=r_{f}}$, then when multiplying the values of the domain, the range must be multiplied by a constant:
${\displaystyle f\left(x_{2}\right)=f\left(x_{1}\right)r_{f}.}$
* Since the domain of ${\displaystyle f}$ must be a geometric sequence multiplied by a common ratio ${\displaystyle r}$, multiply each value of ${\displaystyle x}$ by ${\displaystyle r=2}$. The following geometric sequence is formed: since ${\displaystyle x\cdot 2}$, given ${\displaystyle f(2x)=2{\sqrt {2x}}}$, ${\displaystyle f(2x)}$ will form ${\displaystyle {2{\sqrt {2}},\,4,\,2{\sqrt {6}},\,2{\sqrt {8}},\ldots }}$. Each sequence must have values multiplied by ${\displaystyle {\sqrt {2}}}$. As such, the range of ${\displaystyle f}$ must be geometric for this given ${\displaystyle r}$ value.
• The question is asking whether or not ${\displaystyle f(xr)}$ will give a range for ${\displaystyle f(x)}$ such that you multiply by a constant. Let's define that constant as ${\displaystyle r_{f}}$. Let ${\displaystyle f(x)=f\left(x_{1}\right)}$. Since ${\displaystyle f\left(x_{1}\right)=a\left(x_{1}\right)^{p}}$, if ${\displaystyle x_{2}=x_{1}r}$, then ${\displaystyle f\left(x_{2}\right)=a\left(x_{2}\right)^{p}}$. Since ${\displaystyle x_{2}=x_{1}r}$, function ${\displaystyle f\left(x_{2}\right)=a\left(x_{1}r\right)^{p}}$. Since any term ${\displaystyle (ab)^{m}}$ will give ${\displaystyle a^{m}b^{m}}$, function ${\displaystyle f\left(x_{2}\right)=ax_{1}^{p}r^{p}}$. Since ${\displaystyle ax_{1}^{p}=f\left(x_{1}\right)}$, function ${\displaystyle f\left(x_{2}\right)=f\left(x_{1}\right)r^{p}}$. Since ${\displaystyle r}$ and ${\displaystyle p}$ are constant, the expression ${\displaystyle r^{p}}$ will be constant. Since you are multiplying the range, ${\displaystyle f\left(x_{1}\right)}$, by a common ratio (otherwise known as a geometric sequence), given the constant ${\displaystyle r^{p}=r_{f}}$, then when multiplying the values of the domain, the range must be multiplied by a constant:
${\displaystyle f\left(x_{2}\right)=f\left(x_{1}\right)r_{f}.}$

## Series

It is great to find these patterns to these sequences, but is that the only use we have for these sequences? As always, the answer in math is never no in regards to a low amount of utility (i.e. usefulness). A function does not only describe the pattern associated with numbers but also predicts the graph created when plotted using ${\displaystyle x}$ for its inputs and ${\displaystyle y}$ for its outputs. Using the terms of a sequence, can we determine a sum? We can, and it is called a series.

Definition of a Series.

A series is the sum of the terms of a sequence.

There is always a way to find the sum of a sequence by force: add up the terms one by one and get the answer. Mathematicians are lazy people and don't want to do more work than necessary. That is to say, mathematicians work smart, not hard.

### Arithmetic Series

 Example 2.1.1.a: What is ${\displaystyle {1+2+3+\ldots +97+98+99+100}}$? This is a classic example among many mathematicians. You may be reaching for your calculator, but that will definitely take a long time to punch in all those numbers, and you may accidently make mistakes while writing out the expression. The question is, how do we solve this in the short amount of time we are given on the CLEP exam? Notice that there is a pattern at play: the first term of the series, ${\displaystyle 1}$, plus the last term of the series, ${\displaystyle 100}$, will give us ${\displaystyle 101}$. The second term of the series, ${\displaystyle 2}$, plus the second-to-last term of the series, ${\displaystyle 99}$, will give us ${\displaystyle 101}$. In fact, for all ${\displaystyle 100}$ terms in the series, each pair of numbers, according to their "placement," adds to ${\displaystyle 101}$. Since there are ${\displaystyle 50}$ pairs in this series, and the sum of each pair of numbers always yields ${\displaystyle 101}$, ${\displaystyle 101\cdot 50=5050{\text{ is the sum of the expression above.}}}$

Let us make a conjecture for the above statement. After all, we are not robots that are simply fed a method and do something the same way. We will contract parts of the above expression in the question above to see if our method will be useful for any number of terms. Let ${\displaystyle n}$ be the number of terms in the arithmetic series ${\displaystyle A_{E}}$, let ${\displaystyle S}$ be the sum of ${\displaystyle A_{E}}$, and let ${\displaystyle S_{m}}$ be the column in which the same method we used in Example 2.1.1.a will, hopefully, get us the sum ${\displaystyle S}$.

${\displaystyle {\begin{array}{|c|c|}n&A_{E}&S&S_{m}\\\hline 2&1+2&3&(1+2)\cdot 1=3\\\hline 3&1+2+3&6&(1+3)\cdot 1.5=6\\\hline 4&1+2+3+4&10&(1+4)\cdot 2=10\\\hline 5&1+2+3+4+5&15&(1+5)\cdot 2.5=15\\\hline \vdots &\vdots &\vdots &\vdots \\\hline n&1+2+\ldots +(n-1)+n&S&(1+n)\cdot {n \over 2}=S\end{array}}\!}$

Notice how the odd terms have this weird (or perhaps odd) behavior where not all pairs of terms have the same sum (because there is not an even amount of pairs). While it may seem that using the method in Example 2.1.1.a will work for all odd number of terms based on the table above, maybe it is not true for some really high odd term. Therefore, it is important to prove this is true. For now, let us simply define the formula to be true for both even and odd number of terms.

Gaussian method for finding the sum of an arithmetic series.

For any finite sequence ${\displaystyle {\left\{x_{i}\right\}}}$, where ${\displaystyle x_{i}=a+(i-1)d}$, that contains ${\displaystyle k}$ terms, the sum of each term is

${\displaystyle S_{A}={a+x_{k} \over 2}\cdot k.}$

A finite set of things is something that has a determined number of anythng within that set. An infinite set is a set not described by a determined number of the amount within that set.

Note: the above formula is not known as the "Gaussian method" among mathematicians. This wikibooks will simply refer to this method as the "Gaussian method."[see footnote 1.]

 Example 2.1.1.b: Knowing the sum to an arithmetic series is ${\displaystyle 600}$ and the first term is ${\displaystyle 50}$ but the last term is ${\displaystyle 250}$, what is the constant difference? This seems like an impossible problem to someone who does not know about the Gaussian method or the properties of an arithmetic sequence. However, since you have paid attention, you can figure out for yourself. For the purposes of giving an example, this Wikibooks will explain. We do not know how many terms there are in the sequence, so let's find out how many terms there are. Since ${\displaystyle S_{A}={a+x_{k} \over 2}\cdot k,}$ ${\displaystyle 600={50+250 \over 2}\cdot k.}$ This subject is not called College Algebra for nothing, so do some algebra. ${\displaystyle 600={300 \over 2}\cdot k}$ ${\displaystyle 600=150k}$ ${\displaystyle 40=k}$ Note, however, that we are not looking for the number of terms in the series. We want to know the constant difference of the arithmetic series. Therefore, use the arithmetic sequence direct formula: ${\displaystyle x_{k}=a+d(k-1)}$! ${\displaystyle x_{40}=250}$ ${\displaystyle 250=50+d(40-1)}$ ${\displaystyle 200=39d}$ ${\displaystyle {200 \over 39}=d}$ ${\displaystyle 5{5 \over 39}=d}$

Example 2.2.1.b helps us know how to find the number of terms and the constant difference. Sometimes, you may simply not know one or both of the information you needed to find in Example 2.2.1.b., more often the number of terms. The next example helps illustrate the usefulness of knowing about the properties of arithmetic series and arithmetic sequences.

 Example 2.1.1.c: What is the sum of the arithmetic series ${\displaystyle {4+9+14+\ldots +2,494+2,499+2,504}}$? To find the sum, we need to know how many terms there are; otherwise we cannot use the Gaussian method. Because we do not know the sum of the series, let's use the direct arithmetic sequence formula (since we know which terms correspond to place). We want to find the number of terms, ${\displaystyle k}$, so use the last term (${\displaystyle x_{k}=2,504}$). First, find the constant difference. To get from ${\displaystyle 4}$ to ${\displaystyle 9}$, we need to add ${\displaystyle 5}$ to ${\displaystyle 4}$. Therefore, the constant difference is ${\displaystyle d=5}$. To put it more formally: if recursive formula ${\displaystyle x_{1}=4;x_{2}=4+d}$ and ${\displaystyle x_{2}=9}$, then ${\displaystyle x_{1}=4;9=4+d}$ means ${\displaystyle d=5}$. Second, find the number of terms. If ${\displaystyle x_{k}=2,504}$, ${\displaystyle d=5}$, and ${\displaystyle x_{1}=a=4}$, then ${\displaystyle 2,504=4+5(k-1)}$. Solve for ${\displaystyle k}$: ${\displaystyle 2,500=5(k-1)}$ ${\displaystyle 500=k-1}$ ${\displaystyle 501=k.}$ Finally, use the Gaussian Method: ${\displaystyle S_{A}={4+2,504 \over 2}\cdot 501}$ ${\displaystyle S_{A}={2,508 \over 2}\cdot 501}$ ${\displaystyle S_{A}=1,254\cdot 501}$ ${\displaystyle S_{A}=628,254.}$

As you can see, an arithmetic series can be useful in describing any type of sequence. Now that we have sufficiently explored arithmetic series, let's prove that the Gaussian method works for odd number of terms.

Note: what you are about to learn is NOT required for the curriculum. If you do not understand this proof, do not worry, for it does not matter for the CLEP exam. These proofs are only to build a mathematical understanding of the concepts. As such, you may skip these if you want.

Given that ${\displaystyle {a+x_{k} \over 2}\cdot k}$ finds the sum of an arithmetic series, Prove that ${\displaystyle {a+x_{k} \over 2}\cdot k}$ works for odd number terms.

Given there are ${\displaystyle k}$ terms in the arithmetic sequence, ${\displaystyle x_{k}=a+d(k-1)}$.

Adding one more term makes the sequence even because ${\displaystyle k}$ is odd: ${\displaystyle x_{k+1}=a+d([k+1]-1)=a+dk}$.

Since ${\displaystyle x_{k+1}=a+dk}$, adding ${\displaystyle x_{k+1}}$ to the series makes that the last term, so ${\displaystyle {a+a+dk \over 2}\cdot (k+1)}$.

Because ${\displaystyle {x \over z}=xz^{-1}}$ and ${\displaystyle xz=zx}$, ${\displaystyle 2^{-1}\cdot (a+a+dk)\cdot (k+1)=(a+a+dk)\cdot 2^{-1}(k+1)=(a+a+dk)\cdot {k-1 \over 2}}$.

${\displaystyle -(a+dk)+(a+a+dk)\cdot {k-1 \over 2}}$ will help us find the number of terms we are looking for:

${\displaystyle {{(2a+dk)(k-1) \over 2}-a-dk={2ak+2a+dk+dk^{2} \over 2}-a-dk={2ak+2a+dk+dk^{2} \over 2}-{2a-2dk \over 2}=}}$
${\displaystyle {{2ak+2a+dk+dk^{2}-2a-2dk \over 2}={2ak-dk+dk^{2} \over 2}={k(2a-d+dk) \over 2}={k \over 2}\cdot (2a-d+dk).}}$
Since ${\displaystyle -d+dk}$ has multiplier ${\displaystyle d}$ for both terms, ${\displaystyle d(k-1)}$.
${\displaystyle {[2a+d(k-1)]\cdot {k \over 2}=[a+a+d(k-1)]\cdot {k \over 2}}.}$
Because ${\displaystyle x_{k}=a+d(k-1)}$, ${\displaystyle {[a+a+d(k-1)]\cdot {k \over 2}=\left(a+x_{k}\right)\cdot {k \over 2}={a+x_{k} \over 2}\cdot k}}$.

${\displaystyle {a+x_{k} \over 2}\cdot k}$ is the same formula we started with; ergo, the same formula works for odd number terms and even number terms.

#### CLEP Practice Problems: Check Your Understanding

1 Given the arithmetic series ${\displaystyle 50+45+30+\ldots -700}$, determine the number of terms in the arithmetic series.

 . (Note: typing a comma will give you a decimal, so do not type a comma.)

2 Given the arithmetic series ${\displaystyle 50+45+30+\ldots -700}$, determine the sum of the arithmetic series.

 . (Note: typing a comma will give you a decimal, so do not type a comma.)

3 Function ${\displaystyle S}$ has the following properties: ${\displaystyle S(x)=x{\text{ if }}-20\leq x\leq 50}$ and ${\displaystyle S(x)=3x+1{\text{ if }}50. A student decided to add all terms for all ${\displaystyle x\in \mathbb {Z} }$ or for all x that belong to the set of integers. What is the sum of the terms of function ${\displaystyle S}$ for all ${\displaystyle x\in \mathbb {Z} }$?

 ${\displaystyle 56,625}$ ${\displaystyle 57,641}$ ${\displaystyle 57,690}$ ${\displaystyle 57,450}$ ${\displaystyle 56,776}$

4 The first term of an arithmetic sequence is ${\displaystyle a<0}$. Let there be two separate sequences, ${\displaystyle \left\{x_{i}\right\}=a+d(i-1)}$ and ${\displaystyle \left\{a_{k}\right\}=a+d(k-1)}$, with ${\displaystyle 500}$ terms in each sequence. If the constant difference is ${\displaystyle d<0}$ for ${\displaystyle \left\{x_{i}\right\}}$, but the constant difference is ${\displaystyle d>0}$ for ${\displaystyle \left\{a_{k}\right\}}$, which of the following must be true? Indicate all such statements

 ${\displaystyle x_{1}-x_{2}-\ldots -x_{499}-x_{500}-a_{1}-\ldots -a_{499}-a_{500}=998a}$ ${\displaystyle x_{1}+x_{2}+\ldots +x_{499}+x_{500}+a_{1}+\ldots +a_{499}+a_{500}=-1,000a}$ ${\displaystyle x_{500}<0}$

Note, this practice quiz is unfinished.

#### Proof of Arithmetic Series

Note: what you are about to learn is NOT required for the curriculum. If you do not understand this proof, do not worry, for it does not matter for the CLEP exam. These proofs are only to build a mathematical understanding of the concepts. As such, you may skip these if you want.

### Geometric Series

While it is amazing one can find the sum of an arithmetic sequence, what about the sum of a geometric sequence? Thankfully, mathematicians have found a way to calculate this idea. This theorem (not conjecture, as you will hopefully see with the geometric series proof) is one of many ways mathematicians have found solutions to problems coming from the abstract. As always, it is best to understand the way you can use a formula first before getting the formula introduced. To invoke deeper learning, we want to understand the concept, not reciprocate the formula outloud on a test and say, "I know the answer."

 Example 2.2.1.a: What is the sum of the geometric series ${\displaystyle {1+2+4+8+\ldots +512}}$? Let's define a sum ${\displaystyle S}$ for the geometric sequence: ${\displaystyle {1+2+4+8+\ldots +512=S}}$ Notice that all the terms except ${\displaystyle x_{1}=1}$ has a factor ${\displaystyle 2}$. Therefore, move the first term to the other side by subtracting ${\displaystyle 1}$: ${\displaystyle {2+4+\ldots +512=S-1}}$. Then, factor the two on the left side: ${\displaystyle {2(1+2+4+\ldots +256)=S-1}}$ The key discovery of this exercise is noticing that ${\displaystyle {1+2+4+\ldots +256}}$ is ${\displaystyle S}$ but without ${\displaystyle 512}$, so given that ${\displaystyle {S=1+2+4+8+\ldots +512}}$, ${\displaystyle {1+2+4+\ldots +256}=S-512}$ Solve using elementary algebra: ${\displaystyle {2S-1024=S-1}}$ ${\displaystyle {2S-S=-1+1024}}$ ${\displaystyle {S=1023}}$

Notice how we solved the problem. We said that the sum must equal something positive, so we determined that if it does equal something, we could figure it out in some way without having to do any long calculations. Can we use this method for some general geometric sequence? This how we determine a formula. In your Exploration, you will be tasked with presenting a proof for the formula of the geometric series. We will still give you the formula, but you must prove it yourself in the next exploration. Of course, we will also present another way to prove the geometric series formula in the next section.

Exploration 2-1: Given: a geometric series with first term ${\displaystyle a}$, common ratio ${\displaystyle r}$, and ${\displaystyle k}$ terms.
Prove: The sum of a geometric series ${\displaystyle S={a\left(1-r^{k}\right) \over 1-r}}$.
Given ${\displaystyle k}$ terms, and geometric sequence ${\displaystyle \left\{a_{i}\right\}=ar^{i-1}}$ will have added terms up to ${\displaystyle k}$ in this form: ${\displaystyle S=a+ar+ar^{2}+\ldots +ar^{k-1}}$.

We know that ${\displaystyle S-a}$ will make it so that all terms to the right of the equal sign have factor ${\displaystyle r}$. Ergo,

${\displaystyle S-a=r\left(a+ar+ar^{2}+\ldots +ar^{k-2}\right).}$

Note that ${\displaystyle S-ar^{k-1}=a+ar+ar^{2}+\ldots +ar^{k-2}}$, so divide by ${\displaystyle r}$ and apply the identity to the expression to yield

${\displaystyle {S-a \over r}=S-ar^{k-1}.}$

Subtract ${\displaystyle S}$ and write the expression to the left as a fraction:

${\displaystyle {S-a \over r}-{S \over 1}={S-a-Sr \over r}=-ar^{k-1}.}$

Multiply by ${\displaystyle r}$ and add ${\displaystyle a}$:

${\displaystyle S-Sr=a-ar^{k}.}$

Factor ${\displaystyle S}$ on the left, factor ${\displaystyle a}$ on the right and divide by ${\displaystyle 1-r}$:

${\displaystyle S={a\left(1-r^{k}\right) \over 1-r}.}$
Given ${\displaystyle k}$ terms, and geometric sequence ${\displaystyle \left\{a_{i}\right\}=ar^{i-1}}$ will have added terms up to ${\displaystyle k}$ in this form: ${\displaystyle S=a+ar+ar^{2}+\ldots +ar^{k-1}}$.

We know that ${\displaystyle S-a}$ will make it so that all terms to the right of the equal sign have factor ${\displaystyle r}$. Ergo,

${\displaystyle S-a=r\left(a+ar+ar^{2}+\ldots +ar^{k-2}\right).}$

Note that ${\displaystyle S-ar^{k-1}=a+ar+ar^{2}+\ldots +ar^{k-2}}$, so divide by ${\displaystyle r}$ and apply the identity to the expression to yield

${\displaystyle {S-a \over r}=S-ar^{k-1}.}$

Subtract ${\displaystyle S}$ and write the expression to the left as a fraction:

${\displaystyle {S-a \over r}-{S \over 1}={S-a-Sr \over r}=-ar^{k-1}.}$

Multiply by ${\displaystyle r}$ and add ${\displaystyle a}$:

${\displaystyle S-Sr=a-ar^{k}.}$

Factor ${\displaystyle S}$ on the left, factor ${\displaystyle a}$ on the right and divide by ${\displaystyle 1-r}$:

${\displaystyle S={a\left(1-r^{k}\right) \over 1-r}.}$
Formula of the Geometric Series.

For any finite geometric sequence ${\displaystyle {\left\{x_{i}\right\}}}$, where ${\displaystyle x_{i}=ar^{i-1}}$, that contains ${\displaystyle k}$ terms, the product of each term is

${\displaystyle S_{G}={a\left(r^{k}-1\right) \over r-1}}$

for this Wikibooks[see footnote 2].

#### Proof of Geometric Series

Note: what you are about to learn is NOT required for the curriculum. If you do not understand these concepts, do not worry, for it does not matter for the CLEP exam. These proofs are only to build a mathematical understanding of these concepts. As such, you may skip these if you want.

Given a geometric series with ${\displaystyle k}$ terms and a common ratio ${\displaystyle r}$, prove that the sum of the series, ${\displaystyle S}$, equals ${\displaystyle {a\left(r^{k}-1\right) \over r-1}}$.

By definition, a geometric series has the sequence ${\displaystyle x_{i}=x_{1}\cdot r^{i-1}}$, with ${\displaystyle x_{1}=a}$. As a result, the following is true: ${\displaystyle a+ar+ar^{2}+\ldots +ar^{k-2}+ar^{k-1}}$.

Let ${\displaystyle a+ar+ar^{2}+\ldots +ar^{k-2}+ar^{k-1}=S}$.

Add ${\displaystyle ar^{k}}$ to both sides:

${\displaystyle a+ar+ar^{2}+\ldots +ar^{k-2}+ar^{k-1}+ar^{k}=S+ar^{k}.}$

Subtract ${\displaystyle a}$ to the other side:

${\displaystyle ar+ar^{2}+\ldots +ar^{k-2}+ar^{k-1}+ar^{k}=S+ar^{k}-a.}$

Factor the common ratio ${\displaystyle r}$ to all terms in the left side of the equation:

${\displaystyle r\left(a+ar+ar^{2}+\ldots +ar^{k-2}+ar^{k-1}\right)=S+ar^{k}-a.}$

Factor ${\displaystyle a}$ for the expression ${\displaystyle ar^{k}-a}$ to the right side of the equation:

${\displaystyle r\left(a+ar+ar^{2}+\ldots +ar^{k-2}+ar^{k-1}\right)=S+a\left(r^{k}-1\right).}$

Notice that because ${\displaystyle a+ar+ar^{2}+\ldots +ar^{k-2}+ar^{k-1}=S}$, we can substitute that into the expression on the left side:

${\displaystyle rS=S+a\left(r^{k}-1\right).}$

Finally, solve for ${\displaystyle S}$:

${\displaystyle rS-S=a\left(r^{k}-1\right)}$

${\displaystyle S(r-1)=a\left(r^{k}-1\right)}$

${\displaystyle S={a\left(r^{k}-1\right) \over r-1}.}$

Therefore, the sum of a geometric series must be ${\displaystyle S={a\left(r^{k}-1\right) \over r-1}}$.

### Sigma Notation

How many of you are tired of writing out the expressions? How many of you are tired of figuring out how many expressions there are in a series? Where is this short-hand that we conveniently have for an arithmetic or geometric sequence? Thankfully, all of your answers are coming soon.

Say you want to write out the expression ${\displaystyle \underbrace {1+1+1+1+\ldots +1+1} _{500}}$, in which there are ${\displaystyle 500}$ terms in the series, each of which are the number ${\displaystyle 1}$. We hopefully can evaluate the series without having to think too much about it, or even think about a short-hand about this expression immediately without too much thought. However, the purpose of this easy exercise is to introduce a new notation.

Let ${\displaystyle \sum }$ denote the sum of a series. Located at the bottom of the symbol is the start of the series at index ${\displaystyle i=a}$ (blue) and the top of the symbol is the last index of the series, ${\displaystyle k}$ (red). The term inside the parentheses represents the sequence that the series follows, ${\displaystyle f(i)}$ (orange). The use of the notation of is shown below

${\displaystyle \sum _{\color {Blue}i=a}^{\color {Red}k}\left({\color {Orange}f\left(i\right)}\right).}$
Here, we know that ${\displaystyle \left\{1,1,1,1,\ldots 1,1\right\}}$ must have the sequence ${\displaystyle x_{i}=1+0(i-1)=1=f\left(i\right)}$.

Interpreting the meaning of ${\displaystyle f(i)}$ tells us that for any number in the sequence, the ${\displaystyle i}$th term of the expression is ${\displaystyle 1}$. Knowing that the sequence starts at index ${\displaystyle 1}$, meaning ${\displaystyle i=1}$, and there are ${\displaystyle 500}$ terms, meaning ${\displaystyle k=500}$, we may write the sigma notation as below:

${\displaystyle \sum _{\color {Blue}i=1}^{\color {Red}500}\left({\color {Orange}1}\right).}$

It is important to write the sigma representation having ${\displaystyle f(i)}$ in parenthesis. Often, if you don't have it, you can confuse other terms by accident. It is for this reason that this Wikibooks recommends writing the ${\displaystyle f(i)}$ term in parentheses.

Sigma representation of a general series

For any series that has formula ${\displaystyle f(i)}$, starting from arbitrary point ${\displaystyle i=a}$, and given ${\displaystyle k}$ terms, the sigma ${\displaystyle \left(\sum \right)}$ representation is written as follows:

${\displaystyle \sum _{i=a}^{k}\left(f\left(i\right)\right).}$

Some more ways in which sigma notation is written is often a short-hand, especially when handwriting. This should be used if you want to save time.

• ${\displaystyle \sum _{i=a}^{k}\left(f(i)\right)=\sum _{a\leq i\leq k}\left(f(i)\right)}$
• ${\displaystyle \sum _{i=a}^{\infty }\left(f(i)\right)=\sum _{i\geq a}\left(f(i)\right)}$

Before we jump into our exploration, it may be a good idea to introduce some more expressions that can be rewritten into its sigma form equivalents. Some of the expressions may not even follow from a formula we have seen so far. Nevertheless, let us continue with this concept.

 Example 2.3.1.a: Write the sequence into its sigma form equivalent: ${\displaystyle {\frac {2}{3}}+{\frac {3}{4}}+{\frac {4}{5}}+\cdots +{\frac {70}{71}}+{\frac {71}{72}}}$ A good idea before getting started is to look for any possible patterns. Because you are working with sigma notation, check whether there has to be a pattern involving addition. Look at the numerator of the first two terms. Let us assume we are starting at index ${\displaystyle i=1}$. To get from the first term in the numerator to the second term in the numerator, one simply has to add 1 to the first term. Similarly, to get from the first term in the denominator to the second term in the denominator, one simply has to add 2 to the first term. Both of these are true on a term-by-term basis. ${\displaystyle f(i)={\frac {i+1}{i+2}}}$. This is easy to verify for every term in its respective index. Because this function is true, let us use this to find the number of terms in the expression (i.e., look at the final term in the expression). Let ${\displaystyle f(k)={\frac {71}{72}}={\frac {k+1}{k+2}}}$. There are many ways to solve for ${\displaystyle k}$, the last term of the expression, in this equation. We will show the standard way. ${\displaystyle {\frac {k+1}{k+2}}={\frac {71}{72}}}$ ${\displaystyle \Leftrightarrow 72\cdot (k+1)=71\cdot (k+2)}$ ${\displaystyle \Leftrightarrow 72k+72=71k+142}$ ${\displaystyle \Leftrightarrow k=70}$ We have demonstrated this expression involves 70 terms. With the index, number of terms, and function ready, we may now write the sigma equivalent: ${\displaystyle \sum _{i=1}^{70}\left({\frac {i+1}{i+2}}\right)\blacksquare }$ Alternatively, one can change the index to get a completely different sigma notation. Notice how when the index starts at an ${\displaystyle i>1}$, the formula and the final term's index also changes. Here, we will look specifically at ${\displaystyle i=2}$. Keeping in mind the index represents the placement along the series, going from 2 to 3 is simply a matter of changing along with the numerator. Because the denominator is simply the same idea except displaced by 1 more than the index, the denominator follows a pattern of ${\displaystyle i+1}$. Therefore, the function of the sigma notation is ${\displaystyle g(i)={\frac {i}{i+1}}}$. This is easy to verify for every term in its respective index. Use this function to find the the final term's index. Let ${\displaystyle f(k)={\frac {71}{72}}={\frac {k}{k+1}}}$. We will solve this one the simplest way. Because both the numerator and the denominator are constant, non-zero terms, and ${\displaystyle k}$ is simply an expression of the index for its final term, one can simply set the numerator equal to the other numerator, and the same is true for the denominator. That is, ${\displaystyle {\begin{cases}k=71\\k+1=72\end{cases}}}$ This is perhaps the easiest systems of equations one had to deal with this year learning college algebra. From this, it is obvious ${\displaystyle k=71}$. Knowing the index starting point, function, and final index, the sigma form of this situation is ${\displaystyle \sum _{i=2}^{71}\left({\frac {i}{i+1}}\right)\blacksquare }$ Notice how the pattern changed. Quite intriguing, would you not say?

Some expressions are too hard to determine simply from looking. These are likely going to involve factors multiplied to another. THe best way to determine such a pattern would be to divide terms to see what the factors are.

 Example 2.3.1.b: Write the sequence into its sigma form equivalent: ${\displaystyle 0-2-2+0+4+10+\ldots }$ There seems to be no pattern here. The best thing to do is to determine any similarities between terms and see if there seems to be a pattern from there. Notice how the first and fourth term are zero. This means that there is a term that allows us to obtain a value of zero. Let us assume they are linear functions. If ${\displaystyle k=0}$ is the index of the first term and ${\displaystyle k=3}$ is the third term, then a possible summation function is ${\displaystyle \sum _{k\geq 0}\left(k(k-3)\right)}$. Let us apply this and see if it works. ${\displaystyle k=0\Rightarrow 0}$ ${\displaystyle k=1\Rightarrow -2}$ ${\displaystyle k=2\Rightarrow -2}$ ${\displaystyle k=3\Rightarrow 0}$ ${\displaystyle k=4\Rightarrow 4}$ ${\displaystyle k=5\Rightarrow 10}$ It seems this function works for each term. Therefore, this works. ${\displaystyle \sum _{k=0}^{\infty }\left(k(k-3)\right)=\sum _{k=0}^{\infty }\left(k^{2}-3k\right)\blacksquare }$

More examples will be added later. The next exploration will ask you write out a variety of expressions using sigma form:

Exploration 2-2: Rewrite each expression below into its sigma form equivalent:

(a) ${\displaystyle 1+3+5+7+9+\ldots +541+543}$
(b) ${\displaystyle 1+{1 \over 2}+{1 \over 4}+{1 \over 8}}$
(c) ${\displaystyle 1+2+3+\ldots +99+100+102+104+\ldots +200}$

(d) ${\displaystyle 100-20+4-{4 \over 5}+{4 \over 25}-100-120-140}$
To answer each question, you need to know how many terms are in each series. The only way to determine that is by determining the type of sequence being applied.

(a) Remember how we needed to find out how many terms there are in a sequence and the constant difference to find the sum of an arithmetic sequence? It is the same here. The sigma representation, well, represents the adding of each term in the series. In which case, we need to know the constant difference and the number of terms in the series to determine any validity from it. Note how ${\displaystyle \left\{1,\,3,\,5,\,\ldots ,\,543\right\}}$ is an arithmetic sequence that has the common difference two. Therefore, given the first term is ${\displaystyle 1}$, the arithmetic sequence formula is ${\displaystyle x_{i}=1+2(i-1)=f(i)}$. From this, we can determine the final term's position: ${\displaystyle 543=1+2(i-1)}$. After solving for ${\displaystyle i}$, you find that there are ${\displaystyle i=273}$ terms in the sequence. As such, you may finally write out the final answer:

${\displaystyle \sum _{i=1}^{273}\left(1+2\left(i-1\right)\right).}$

Note, because you can start counting the terms at ${\displaystyle i=0}$, you may rewrite the expression as follows:
${\displaystyle \sum _{i=0}^{272}\left(1+2i\right).}$

(b) Use the same idea from item (a) to rewrite the expression into its sigma equivalent. Notice that the series shows a geometric series, with common ratio of ${\displaystyle {1 \over 2}}$ starting at ${\displaystyle 1}$. Therefore, ${\displaystyle f(i)=\left({1 \over 2}\right)^{i-1}}$. Since there is no need to use a formula to find out how many terms there are in the series, simply count and you have your answer:

${\displaystyle \sum _{i=1}^{4}\left({1 \over 2}\right)^{i-1}.}$

Again, you may rewrite the sigma form equivalent starting at ${\displaystyle i=0}$:
${\displaystyle \sum _{i=0}^{3}\left({1 \over 2}\right)^{i}.}$

(c) There are two series shown in the expression. Separate each series within as the following: ${\displaystyle f_{1}\left(i\right)=1+2+3+\ldots +99}$ and ${\displaystyle f_{2}\left(i\right)=100+102+104+\ldots +200}$. Both ${\displaystyle f_{1}}$ and ${\displaystyle f_{2}}$ are arithmetic sequences but with different differences and starting numbers. Instead of explaining each one, only the formula will be provided for each sequence: ${\displaystyle f_{1}(i)=1+1(i-1)}$ and ${\displaystyle f_{2}(i)=100+2(i-1)}$. Note that the starting number and end numbers are not appropriate to use in the same sigma notation, so separate them into two sigma representations and add them together. Thus, the sigma representation for this expression is

${\displaystyle \sum _{i=1}^{99}(1+1(i-1))+\sum _{i=1}^{51}(100+2(i-1)).}$

Do you see why we ask you to write the formula using parentheses? Also, you may have separated the series differently. In which case, if ${\displaystyle f_{1}\left(i\right)=1+2+3+\ldots +99+100}$ and ${\displaystyle f_{2}\left(i\right)=102+104+\ldots +200}$, then ${\displaystyle f_{1}(i)=1+1(i-1)}$ and ${\displaystyle f_{2}(i)=102+2(i-1)}$, where ${\displaystyle f_{1}}$ has ${\displaystyle 100}$ terms and ${\displaystyle f_{1}}$ has ${\displaystyle 50}$ terms. Therefore, another representation is also appropriate:
${\displaystyle \sum _{i=1}^{100}(1+1(i-1))+\sum _{i=1}^{50}(100+2(i-1)).}$

(d) There are two series shown in the expression. Separate each series within as the following: ${\displaystyle f_{1}\left(i\right)=100-20+4-{4 \over 5}+{4 \over 25}}$ and ${\displaystyle f_{2}\left(i\right)=-100-120-140}$. While ${\displaystyle f_{1}}$ is geometric, with common ratio ${\displaystyle -{1 \over 5}}$, ${\displaystyle f_{2}}$ is arithmetic, with constant difference ${\displaystyle -20}$. Instead of explaining each one, only the formula will be provided for each sequence: ${\displaystyle f_{1}(i)=100\left(-{1 \over 5}\right)^{i-1}}$ and ${\displaystyle f_{2}(i)=-100-20(i-1)}$. Because of the difference in formula and difference in starting numbers used, it is impossible to put them together in the same sum. Therefore, the sigma form of the series is

${\displaystyle \sum _{i=1}^{5}\left(100\left(-{1 \over 5}\right)^{i-1}\right)+\sum _{i=1}^{3}(-100-20(i-1)).}$

Notice how you added the other term instead of substracting. If you were to subtract the second sigma representation you would in effect flip the sign on every other term. As such, the second series representation would be wrong unless you were to change it to this:

${\displaystyle \sum _{i=1}^{5}\left(100\left(-{1 \over 5}\right)^{i-1}\right)-\sum _{i=1}^{3}(100+20(i-1)).}$
To answer each question, you need to know how many terms are in each series. The only way to determine that is by determining the type of sequence being applied.

(a) Remember how we needed to find out how many terms there are in a sequence and the constant difference to find the sum of an arithmetic sequence? It is the same here. The sigma representation, well, represents the adding of each term in the series. In which case, we need to know the constant difference and the number of terms in the series to determine any validity from it. Note how ${\displaystyle \left\{1,\,3,\,5,\,\ldots ,\,543\right\}}$ is an arithmetic sequence that has the common difference two. Therefore, given the first term is ${\displaystyle 1}$, the arithmetic sequence formula is ${\displaystyle x_{i}=1+2(i-1)=f(i)}$. From this, we can determine the final term's position: ${\displaystyle 543=1+2(i-1)}$. After solving for ${\displaystyle i}$, you find that there are ${\displaystyle i=273}$ terms in the sequence. As such, you may finally write out the final answer:

${\displaystyle \sum _{i=1}^{273}\left(1+2\left(i-1\right)\right).}$

Note, because you can start counting the terms at ${\displaystyle i=0}$, you may rewrite the expression as follows:
${\displaystyle \sum _{i=0}^{272}\left(1+2i\right).}$

(b) Use the same idea from item (a) to rewrite the expression into its sigma equivalent. Notice that the series shows a geometric series, with common ratio of ${\displaystyle {1 \over 2}}$ starting at ${\displaystyle 1}$. Therefore, ${\displaystyle f(i)=\left({1 \over 2}\right)^{i-1}}$. Since there is no need to use a formula to find out how many terms there are in the series, simply count and you have your answer:

${\displaystyle \sum _{i=1}^{4}\left({1 \over 2}\right)^{i-1}.}$

Again, you may rewrite the sigma form equivalent starting at ${\displaystyle i=0}$:
${\displaystyle \sum _{i=0}^{3}\left({1 \over 2}\right)^{i}.}$

(c) There are two series shown in the expression. Separate each series within as the following: ${\displaystyle f_{1}\left(i\right)=1+2+3+\ldots +99}$ and ${\displaystyle f_{2}\left(i\right)=100+102+104+\ldots +200}$. Both ${\displaystyle f_{1}}$ and ${\displaystyle f_{2}}$ are arithmetic sequences but with different differences and starting numbers. Instead of explaining each one, only the formula will be provided for each sequence: ${\displaystyle f_{1}(i)=1+1(i-1)}$ and ${\displaystyle f_{2}(i)=100+2(i-1)}$. Note that the starting number and end numbers are not appropriate to use in the same sigma notation, so separate them into two sigma representations and add them together. Thus, the sigma representation for this expression is

${\displaystyle \sum _{i=1}^{99}(1+1(i-1))+\sum _{i=1}^{51}(100+2(i-1)).}$

Do you see why we ask you to write the formula using parentheses? Also, you may have separated the series differently. In which case, if ${\displaystyle f_{1}\left(i\right)=1+2+3+\ldots +99+100}$ and ${\displaystyle f_{2}\left(i\right)=102+104+\ldots +200}$, then ${\displaystyle f_{1}(i)=1+1(i-1)}$ and ${\displaystyle f_{2}(i)=102+2(i-1)}$, where ${\displaystyle f_{1}}$ has ${\displaystyle 100}$ terms and ${\displaystyle f_{1}}$ has ${\displaystyle 50}$ terms. Therefore, another representation is also appropriate:
${\displaystyle \sum _{i=1}^{100}(1+1(i-1))+\sum _{i=1}^{50}(100+2(i-1)).}$

(d) There are two series shown in the expression. Separate each series within as the following: ${\displaystyle f_{1}\left(i\right)=100-20+4-{4 \over 5}+{4 \over 25}}$ and ${\displaystyle f_{2}\left(i\right)=-100-120-140}$. While ${\displaystyle f_{1}}$ is geometric, with common ratio ${\displaystyle -{1 \over 5}}$, ${\displaystyle f_{2}}$ is arithmetic, with constant difference ${\displaystyle -20}$. Instead of explaining each one, only the formula will be provided for each sequence: ${\displaystyle f_{1}(i)=100\left(-{1 \over 5}\right)^{i-1}}$ and ${\displaystyle f_{2}(i)=-100-20(i-1)}$. Because of the difference in formula and difference in starting numbers used, it is impossible to put them together in the same sum. Therefore, the sigma form of the series is

${\displaystyle \sum _{i=1}^{5}\left(100\left(-{1 \over 5}\right)^{i-1}\right)+\sum _{i=1}^{3}(-100-20(i-1)).}$

Notice how you added the other term instead of substracting. If you were to subtract the second sigma representation you would in effect flip the sign on every other term. As such, the second series representation would be wrong unless you were to change it to this:

${\displaystyle \sum _{i=1}^{5}\left(100\left(-{1 \over 5}\right)^{i-1}\right)-\sum _{i=1}^{3}(100+20(i-1)).}$

#### Sigma Simplification Techniques

In some textbooks, this section would be called rules. We call them what they really are: techniques to simplify common sigma representations.

For any series that has constant term ${\displaystyle c}$, starting from index ${\displaystyle i=1}$, and given ${\displaystyle k}$ terms,

${\displaystyle \sum _{i=1}^{k}\left(c\right)=ck.}$

As always, if you are confused, write out some terms. The above sigma representation states that for any constant term ${\displaystyle c}$ added ${\displaystyle k}$ times, the resulting sum is equivalent to ${\displaystyle \underbrace {c+c+c+\ldots +c+c+c} _{k}=ck}$. Never forget that multiplication is repeated addition. This axiom you heard in elementary school is still important to this day.

For any series that has constant term ${\displaystyle c}$ multiplied by changing index ${\displaystyle i}$, starting from index ${\displaystyle i=1}$, and given ${\displaystyle k}$ total terms,

${\displaystyle \sum _{i=1}^{k}\left(ci\right)=c\sum _{i=1}^{k}\left(i\right).}$

This second "rule" can be simplified as the following. Since ${\displaystyle \sum _{i=1}^{k}\left(ci\right)}$, we can conclude that ${\displaystyle \sum _{i=1}^{k}\left(ci\right)=1c+2c+3c+\ldots +c(k-2)+c(k-1)+ck}$. Each term in the expression ${\displaystyle 1c+2c+3c+\ldots +c(k-2)+c(k-1)+ck}$ has common factor ${\displaystyle c}$, so ${\displaystyle c(1+2+3+\ldots +(k-2)+(k-1)+k)}$. Notice that each term on the inside of the parenthesis is a basic series where ${\displaystyle i}$ starting at ${\displaystyle 1}$ goes to final term ${\displaystyle k}$ as a sum, so

${\displaystyle \sum _{i=1}^{k}\left(ci\right)=c\sum _{i=1}^{k}\left(i\right).}$

For any series that has constant term ${\displaystyle c}$ in addition to changing index ${\displaystyle i}$, starting from index ${\displaystyle i=1}$, and given ${\displaystyle k}$ total terms,

${\displaystyle \sum _{i=1}^{k}\left(c+i\right)=kc+\sum _{i=1}^{k}\left(i\right).}$

This one is harder to see why it is true. As always, however, try it out by hand. Remember that sigma notations are short hands of sums that have some sort of formula, so write out a term in case you are ever confused. Notice that ${\displaystyle \sum _{i=1}^{k}\left(c+i\right)=(c+1)+(c+2)+(c+3)+(c+4)+\ldots +(c+k-1)+(c+k)}$. Because adding is commutative, and because the parentheses do not change the sum of the series, you may group terms such that ${\displaystyle (c+1)+(c+2)+(c+3)+(c+4)+\ldots +(c+k-1)+(c+k)=\underbrace {c+c+c+c+\ldots +c+c} _{k}+(1+2+3+\ldots +k-1+k)}$ We have already determined the sigma notation for each grouping, so we may put it together to say the following is true:

${\displaystyle \sum _{i=1}^{k}\left(c+i\right)=kc+\sum _{i=1}^{k}\left(i\right).}$

For any series that has constant term ${\displaystyle c}$ in addition to the range of the function of ${\displaystyle i}$ or ${\displaystyle f\left(i\right)}$, starting from index ${\displaystyle i=1}$, and given ${\displaystyle k}$ total terms,

${\displaystyle \sum _{i=1}^{k}\left(c+f\left(i\right)\right)=kc+\sum _{i=1}^{k}\left(f\left(i\right)\right).}$

Realize that the above "rule" is simply an extension of the previous one. If ${\displaystyle f(i)=i}$, then ${\displaystyle \sum _{i=1}^{k}\left(c+f\left(i\right)\right)=kc+\sum _{i=1}^{k}\left(f\left(i\right)\right)}$. Realize that any function can comply with this rule. By extension:

${\displaystyle \sum _{i=1}^{k}\left(cf\left(i\right)\right)=c\sum _{i=1}^{k}\left(f\left(i\right)\right)}$

Using this general knowledge, one can also argue the next following "rule":

For any series that has constant term ${\displaystyle a}$ multiplied by the function of ${\displaystyle i}$ or ${\displaystyle f\left(i\right)}$, starting from index ${\displaystyle i=1}$, when adding constant ${\displaystyle c}$ to the product, and given ${\displaystyle k}$ total terms,

${\displaystyle \sum _{i=1}^{k}\left(c+af\left(i\right)\right)=kc+a\sum _{i=1}^{k}\left(f\left(i\right)\right).}$

Given the way we found the same identities, you may do the proof for this one by yourself as an exercise in sigma notation. The best way to understand this new notation is to practice it. (Of course, this Wikibooks will provide plenty of practice.)

Along with that practice, you may also try to show that the identity for the next "rule" is also true:

For any series in which the formula is two different arbitrary functions: ${\displaystyle f\left(i\right)}$ and ${\displaystyle g\left(i\right)}$, starting from index ${\displaystyle i=1}$, and given ${\displaystyle k}$ total terms,

${\displaystyle \sum _{i=1}^{k}\left(f\left(i\right)+g\left(i\right)\right)=\sum _{i=1}^{k}\left(f\left(i\right)\right)+\sum _{i=1}^{k}\left(g\left(i\right)\right).}$

With this, you now have a sufficient foundation for the necessary tools needed to prove and disprove statements as well as create your own identity. Before we move on the next section, we must mention the sigma identity for each of the series type we learned in this wikibooks.

#### Sigma Identities for Series Types

What is the sigma identity of an arithmetic series? Before we give you the identity, it is important to understand arithmetic series. If the person reading this wikibooks has jumped from one section to the other, we may recommend you read anything from the previous if the current reasoning seems to not make sense. First, how do we write an arithmetic sequence using a formula? Like this:

${\displaystyle x_{k}=a+(k-1)d.}$

The formula for how to find an arithmetic series is this:
${\displaystyle S_{a}={a+x_{k} \over 2}k.}$

Notice how ${\displaystyle x_{k}}$ is in both formulas, so substitute ${\displaystyle x_{k}}$ for ${\displaystyle a+(k-1)d}$ and you get:
${\displaystyle S_{a}={2a+(k-1)d \over 2}k=ak+{(k-1)d \over 2}k=\left[2a+(k-1)d\right]{k \over 2}.}$

Because ${\displaystyle x_{k}=a+(k-1)d.}$ is also a formula that helps us find each term in the series, we have finally created our sigma identity:

Formula for an Arithmetic Series in Sigma Form

An arithmetic series in which the formula is ${\displaystyle f(i)=a+(i-1)d}$, starting from index ${\displaystyle i=1}$, and given ${\displaystyle k}$ total terms,

${\displaystyle \sum _{i=1}^{k}\left(a+(i-1)d\right)={2a+(k-1)d \over 2}\cdot k.}$

We can use the same process to find the sum of a geometric series. You may consider this your next exercise proving that the sigma form of the geometric series is the same one shown below.

Formula for a Geometric Series in Sigma Form

A geometric series in which the formula is ${\displaystyle f(i)=ar^{i-1}}$, starting from index ${\displaystyle i=1}$, and given ${\displaystyle k}$ total terms,

${\displaystyle \sum _{i=1}^{k}\left(ar^{i-1}\right)={a\left(r^{k}-1\right) \over r-1}.}$

The problems below are basically comprehension questions. If you can do all the problems below, you understand all you need to know about sigma notations. Keep in mind, these problems are much harder than usual CLEP problems; however, being able to do these problems with proficiency proves understanding of the material at a deeper level, which means you are better prepared for easier problems and harder problems down the line.

Non-calculator problem. Expected completion time: 30 minutes.

1. Find the smallest ${\displaystyle x}$ necessary to give the smallest real number for the expression below. After, simplify and calculate for the smallest real value. Do not round your answer, and do not use a calculator or program.

${\displaystyle \log _{5}\left(\sum _{i=1}^{x}\left(-500+{1 \over 3}i\right)\right)}$

Remember, ${\displaystyle \log _{a}c=b}$ is the inverse of an exponential equation ${\displaystyle a^{b}=c}$. The properties are the following: ${\displaystyle \left\{a\vert a>0,\,a\neq 1\right\},\,\left\{b\vert b>0,\,b\in \mathbb {R} \right\}{\text{ and}},\,c>0}$. The real number requirement comes in the form of ${\displaystyle c>0}$, that is the inside of a logarithm must have a result of greater than zero (it cannot equal zero). This is the first step to solving the problem.

${\displaystyle \sum _{i=1}^{x}\left(-500+{1 \over 3}i\right)>0}$

Separate the two expressions inside the sum by applying a property rule, given constant ${\displaystyle C=-500}$ and ${\displaystyle f(i)={1 \over 3}i}$:
${\displaystyle \sum _{i=1}^{x}\left(-500+{1 \over 3}i\right)=-500x+\sum _{i=1}^{x}{1 \over 3}i>0}$

Since ${\displaystyle \sum _{k=1}^{n}Cf(i)=C\sum _{k=1}^{n}f(i)}$, and ${\displaystyle f(i)}$ is arithmetic, it must be that case that
${\displaystyle \sum _{i=1}^{x}{1 \over 3}i={1 \over 3}\cdot {1+x \over 2}x={1 \over 3}\cdot {x(x+1) \over 2}={x(x+1) \over 6}.}$

From there, put the sum of the two expressions over the same denominator and simplify:

${\displaystyle -500x+{x(x+1) \over 6}>0}$
${\displaystyle \Rightarrow {-3000x \over 6}+{x(x+1) \over 6}>0}$
${\displaystyle \Rightarrow {-3000x+x^{2}+x \over 6}>0}$
${\displaystyle \Rightarrow {-2999x+x^{2} \over 6}>0}$
${\displaystyle \Rightarrow -2999x+x^{2}>0}$
${\displaystyle \Rightarrow x(x-2999)>0}$

Finally, note that the zero factor property gives ${\displaystyle ab=0}$ if and only if ${\displaystyle a=0}$ or ${\displaystyle b=0}$ or both. However, for a zero factor property in which the two multipliers give a positive product, it must be the case that either ${\displaystyle a}$ and ${\displaystyle b}$ are positive or ${\displaystyle a}$ and ${\displaystyle b}$ are negative. Through that logic, we can determine four possible solutions: ${\displaystyle x>0}$ and ${\displaystyle x>2999}$, OR ${\displaystyle x<0}$ and ${\displaystyle x<2999}$. Despite the four solutions, only one solution must be true: ${\displaystyle x>2999}$ because sigma notations only work through ${\displaystyle x\in \mathbb {Z^{+}} }$ or ${\displaystyle x}$ being the set of positive integers in the situation described in the problem. As such, all ${\displaystyle x}$ greater than ${\displaystyle 2999}$ gives an acceptable solution. However, this is only the first part of the problem. Now comes the easy part.

The smallest number must be ${\displaystyle x=3000}$ in the situation described above:

{\displaystyle {\begin{aligned}\log _{5}\left(\sum _{i=1}^{3,000}\left(-500+{1 \over 3}i\right)\right)&=\log _{5}\left(-500(3,000)+{3,000(3,000+1) \over 6}\right)\\&=\log _{5}\left(-1,500,000+500(3000+1)\right)\\&=\log _{5}\left(-1,500,000+1,500,000+500)\right)\\&=\log _{5}\left(500\right)\end{aligned}}}

Finally, simplify ${\displaystyle \log _{5}\left(500\right)}$ to get the final answer:

{\displaystyle {\begin{aligned}\log _{5}\left(500\right)&=\log _{5}\left(100\cdot 5\right)\\&=\log _{5}\left(10^{2}\cdot 5\right)\\&=\log _{5}\left((5\cdot 2)^{2}\cdot 5\right)\\&=\log _{5}\left(5^{2}\cdot 2^{2}\cdot 5\right)\\&=\log _{5}\left(2^{2}\cdot 5^{3}\right)\\&=3\log _{5}\left(5\right)+2\log _{5}\left(2\right)+1\\&=3+2\log _{5}\left(2\right)\blacksquare \end{aligned}}}
Remember, ${\displaystyle \log _{a}c=b}$ is the inverse of an exponential equation ${\displaystyle a^{b}=c}$. The properties are the following: ${\displaystyle \left\{a\vert a>0,\,a\neq 1\right\},\,\left\{b\vert b>0,\,b\in \mathbb {R} \right\}{\text{ and}},\,c>0}$. The real number requirement comes in the form of ${\displaystyle c>0}$, that is the inside of a logarithm must have a result of greater than zero (it cannot equal zero). This is the first step to solving the problem.

${\displaystyle \sum _{i=1}^{x}\left(-500+{1 \over 3}i\right)>0}$

Separate the two expressions inside the sum by applying a property rule, given constant ${\displaystyle C=-500}$ and ${\displaystyle f(i)={1 \over 3}i}$:
${\displaystyle \sum _{i=1}^{x}\left(-500+{1 \over 3}i\right)=-500x+\sum _{i=1}^{x}{1 \over 3}i>0}$

Since ${\displaystyle \sum _{k=1}^{n}Cf(i)=C\sum _{k=1}^{n}f(i)}$, and ${\displaystyle f(i)}$ is arithmetic, it must be that case that
${\displaystyle \sum _{i=1}^{x}{1 \over 3}i={1 \over 3}\cdot {1+x \over 2}x={1 \over 3}\cdot {x(x+1) \over 2}={x(x+1) \over 6}.}$

From there, put the sum of the two expressions over the same denominator and simplify:

${\displaystyle -500x+{x(x+1) \over 6}>0}$
${\displaystyle \Rightarrow {-3000x \over 6}+{x(x+1) \over 6}>0}$
${\displaystyle \Rightarrow {-3000x+x^{2}+x \over 6}>0}$
${\displaystyle \Rightarrow {-2999x+x^{2} \over 6}>0}$
${\displaystyle \Rightarrow -2999x+x^{2}>0}$
${\displaystyle \Rightarrow x(x-2999)>0}$

Finally, note that the zero factor property gives ${\displaystyle ab=0}$ if and only if ${\displaystyle a=0}$ or ${\displaystyle b=0}$ or both. However, for a zero factor property in which the two multipliers give a positive product, it must be the case that either ${\displaystyle a}$ and ${\displaystyle b}$ are positive or ${\displaystyle a}$ and ${\displaystyle b}$ are negative. Through that logic, we can determine four possible solutions: ${\displaystyle x>0}$ and ${\displaystyle x>2999}$, OR ${\displaystyle x<0}$ and ${\displaystyle x<2999}$. Despite the four solutions, only one solution must be true: ${\displaystyle x>2999}$ because sigma notations only work through ${\displaystyle x\in \mathbb {Z^{+}} }$ or ${\displaystyle x}$ being the set of positive integers in the situation described in the problem. As such, all ${\displaystyle x}$ greater than ${\displaystyle 2999}$