# Algebra/Binomial Theorem

 Algebra ← Quadratic Equation Binomial Theorem Complex Numbers →

The notation ' $n!$ ' is defined as n factorial.

$n!=n\times (n-1)\times (n-2)\times (n-3)\times \dots \times 3\times 2\times 1$ 0 factorial is equal to 1.

$0!=1$ Proof of 0 factorial = 1

$n!=n\times (n-1)!$ When n = 1,
$1!=1\times (1-1)!$ $1=1\times 0!$ And thus,
$0!=1$ The binomial thereom gives the coefficients of the polynomial

$(x+y)^{n}$ .

We may consider without loss of generality the polynomial, of order n, of a single variable z. Assuming $x\neq 0$ set z = y / x

$(x+y)^{n}=x^{n}(1+z)^{n}$ .

The expansion coefficients of $(1+z)^{n}$ are known as the binomial coefficients, and are denoted

$(1+z)^{n}=\sum _{k=0}^{n}{n \choose k}z^{k}$ .

Noting that

$(x+y)^{n}=\sum _{k=0}^{n}{n \choose k}x^{n-k}y^{k}$ is symmetric in x and y, the identity

${n \choose n-k}={n \choose k}$ may be shown by replacing k by n - k and reversing the order of summation.

A recursive relationship between the ${n \choose k}$ may be established by considering

$(1+z)^{n+1}=(1+z)(1+z)^{n}=\sum _{k=0}^{n+1}{n+1 \choose k}z^{k}=(1+z)\sum _{k=0}^{n}{n \choose k}z^{k}$ or

$\sum _{k=0}^{n+1}{n+1 \choose k}z^{k}=\sum _{k=0}^{n}{n \choose k}z^{k}+\sum _{k=0}^{n}{n \choose k}z^{k+1}=\sum _{k=0}^{n}{n \choose k}z^{k}+\sum _{k=1}^{n+1}{n \choose k-1}z^{k}$ .

Since this must hold for all values of z, the coefficients of $z^{k}$ on both sides of the equation must be equal

${n+1 \choose k}={n \choose k}+{n \choose k-1}$ for k ranging from 1 through n, and

${n+1 \choose n+1}={n \choose n}={\frac {n!}{(n-n)!n!}}={\frac {n!}{n!}}=1$ ${n+1 \choose 0}={n \choose 0}={\frac {n!}{(n-0)!0!}}={\frac {n!}{n!}}=1$ .

Pascal's Triangle is a schematic representation of the above recursion relation ...

Show

${n \choose k}={\frac {n!}{k!(n-k)!}}$ (proof by induction on n).

A useful identity results by setting $z=1$ $\sum _{k=0}^{n}{n \choose k}=2^{n}$ .

## The visual way to do the binomial theorem

(this section is from difference triangles)

Lets look at the results for (x+1)n where n ranges from 0 to 3.

(x+1)0 =          1x0           =                1
(x+1)1 =        1x1+1x0         =              1   1
(x+1)2 =      1x2+2x1+1x0       =            1   2   1
(x+1)3 =    1x3+3x2+3x1+1x0     =          1   3   3   1


This new triangle is Pascal’s Triangle.

It follows a counting method different from difference triangles.

The sum of the x-th number in the n-th difference  and
the (x+1)-th number in the n-th difference yields the
(x+1)-th number in the (n-1)-th difference.


It would take a lot of adding if we were to use the difference triangles in the X-gon to compute (x+1)10. However, using the Pascal’s Triangle which we have derived from it, the task becomes much simpler. Let’s expand Pascal’s Triangle.

(x+1)0                                    1
(x+1)1                                  1   1
(x+1)2                                1   2   1
(x+1)3                             1    3   3    1
(x+1)4                           1    4   6   4    1
(x+1)5                         1   5   10   10   5   1
(x+1)6                      1   6   15   20   15   6   1
(x+1)7                   1   7   21   35   35   21   7    1
(x+1)8                1   8   28   56   70   56   28   8    1
(x+1)9              1   9   36   84   126  126  84   36   9    1
(x+1)10          1   10  45   120  210  252  210  120  45   10    1


The final line of the triangle tells us that

(x+1)10 = 1x10 + 10x9 + 45x8 + 120x7 + 210x6 + 252x5 + 210x4 + 120x3 + 45x2 + 10x1 + 1x0.