Proving the Sum of Cubes Formula using Mathematical Induction

[efekwulsemmay]

Nevermind, got the answer :D

Homework Statement

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}

Homework Equations

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

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}

 

[whs]

<br /> \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}<br />

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

 

[emyt]

Homework Statement

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}

Homework Equations

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

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}

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

 

[truth is life]

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