Induction is a form of proof useful for proving equations involving non-closed expressions (i.e., expressions with
n
{\displaystyle n}
terms; sequences).
Induction involves first proving that the equation is true for
n
=
1
{\displaystyle n=1}
, then proving true for
n
=
k
+
1
{\displaystyle n=k+1}
(assuming for the purpose of the proof that the equation holds true for
n
=
k
{\displaystyle n=k}
). Since it is true for
n
=
k
{\displaystyle n=k}
and true for
n
=
k
+
1
{\displaystyle n=k+1}
, and also true for
n
=
1
{\displaystyle n=1}
, it is true for
n
=
2
{\displaystyle n=2}
. It follows that it is true for all positive integers
n
{\displaystyle n}
.
Q: Prove by mathematical induction that for all integers
n
≥
1
{\displaystyle n\geq 1}
,
1
3
+
2
3
+
3
3
+
4
3
+
⋯
+
n
3
=
(
1
+
2
+
3
+
.
.
.
.
n
)
2
{\displaystyle 1^{3}+2^{3}+3^{3}+4^{3}+\cdots +n^{3}=(1+2+3+....n)^{2}}
A:
When
n
=
1
{\displaystyle n=1}
,
1
3
=
1
=
1
4
(
1
)
2
(
(
1
)
+
1
)
2
=
1
4
(
4
)
=
1
{\displaystyle 1^{3}=1={\tfrac {1}{4}}(1)^{2}((1)+1)^{2}={\tfrac {1}{4}}(4)=1}
, so it is true for
n
=
1
{\displaystyle n=1}
Suppose that the statement is true for
k
,
k
∈
N
{\displaystyle k,k\in \mathbb {N} }
. That is, suppose that
1
3
+
2
3
+
3
3
+
4
3
+
⋯
+
k
3
=
1
4
k
2
(
k
+
1
)
2
{\displaystyle 1^{3}+2^{3}+3^{3}+4^{3}+\cdots +k^{3}={\tfrac {1}{4}}k^{2}(k+1)^{2}}
. This is sometimes called the induction hypothesis .
Then prove the statement for
n
=
k
+
1
{\displaystyle n=k+1}
(that is, prove that
1
3
+
2
3
+
3
3
+
⋯
+
(
k
+
1
)
3
=
1
4
(
k
+
1
)
2
(
k
+
2
)
2
{\displaystyle 1^{3}+2^{3}+3^{3}+\cdots +(k+1)^{3}={\tfrac {1}{4}}(k+1)^{2}(k+2)^{2}}
:
LHS
=
1
3
+
2
3
+
3
3
+
4
+
3
+
⋯
+
k
3
+
(
k
+
1
)
3
=
1
4
k
2
(
k
+
1
)
2
+
(
k
+
1
)
3
(by the induction hypothesis)
=
1
4
(
k
+
1
)
2
(
k
2
+
4
(
k
+
1
)
)
=
1
4
(
k
+
1
)
2
(
k
2
+
4
k
+
4
)
=
1
4
(
k
+
1
)
2
(
k
+
2
)
2
=
RHS
{\displaystyle {\begin{aligned}{\mbox{LHS}}&=1^{3}+2^{3}+3^{3}+4+3+\cdots +k^{3}+(k+1)^{3}\\&={\tfrac {1}{4}}k^{2}(k+1)^{2}+(k+1)^{3}&{\mbox{ (by the induction hypothesis)}}\\&={\tfrac {1}{4}}(k+1)^{2}(k^{2}+4(k+1))\\&={\tfrac {1}{4}}(k+1)^{2}(k^{2}+4k+4)\\&={\tfrac {1}{4}}(k+1)^{2}(k+2)^{2}\\&={\mbox{RHS}}\end{aligned}}}
It follows from parts 1 and 2 by mathematical induction that the statement is true for all positive integers
n
{\displaystyle n}
.