# Proof by Induction Problem

1. Oct 23, 2009

### efekwulsemmay

1. The problem statement, all variables and given/known data
I need to prove using mathematical induction that:

$$\sum^{n}_{i=1} i^{3}= 1^{3}+2^{3}+3^{3}+...+n^{3}=\left[\frac{(n+1)\cdot n}{2}\right]^{2}$$

2. Relevant equations

$$\sum^{n}_{i=1} i^{3}= 1^{3}+2^{3}+3^{3}+...+n^{3}=\left[\frac{(n+1)\cdot n}{2}\right]^{2}$$

3. The attempt at a solution

Ok, I have proved that the statement is true when n=1 and have gone on to assume that it must be true for any arbitrary value of n. I chose n=p. Thus:

$$\sum^{p}_{i=1} i^{3}= 1^{3}+2^{3}+3^{3}+...+p ^{3}=\left[\frac{(p+1)\cdot p}{2}\right]^{2}$$

Where i am stuck is that I know i need to show that if the statement is true when n=p then it must be true for n=p+1. I have gotten as far as:

$$\sum^{p}_{i=1} i^{3}= 1^{3}+2^{3}+3^{3}+...+p^{3}+(p+1)^{3}=\left[\frac{(p+1)\cdot p}{2}\right]^{2}+(p+1)^{3}$$

Last edited: Oct 23, 2009
2. Oct 23, 2009

### whs

$$\sum^{p}_{i=1} i^{3}= 1^{3}+2^{3}+3^{3}+...+p^{3}+(p+1)^{3}=\left[\frac{(p+1)\cdot p}{2}\right]^{2}+(p+1)^{3}$$

Factor out a (p+1)^2, and see what you see.

3. Oct 23, 2009

### emyt

Re: Nevermind, got the answer :D

have you tried just expanding it out and seeing what you get?
the induction step is simple, you got that - the rest is just algebra

4. Oct 23, 2009

### truth is life

Basically, you should expand, collect terms, and reduce. You should be able to do it that way.