Inductive Proof on Well Known Sum

[Shackman]

Homework Statement

Use induction to prove the equation
(1+2+...+n)^2 = 1^3 + 2^3 + ... + n^3

2. The attempt at a solution
I've done three inductive proofs previous to this one where I showed that the equation was true for some case (usually n=1), then assumed it was true for n, and proved it was true for n+1. This proof doesn't seem to fit that form though because for the other proofs I could write it as the right hand side of the equation (since it's assumed that it is true for n) plus the right hand side of the equation when n+1 is plugged in. But in this case..

(1+2+...+n+(n+1))^2 != (1+2+...+n)^2 + (n+1)^2

So, I guess I haven't done a proof like this and could use a nudge in the right direction..

 

[Mystic998]

What is $$\sum^{n}_{i=1} i$$?

 

[Shackman]

n(n+1)/2

In an effort to see where you're going with this, I thought I'd do the same for i^2 and got

n^2(n^2 + 1)/4

Either this isn't correct or I'm not sure where you're headed.

 

[Mystic998]

n(n+1)/2

In an effort to see where you're going with this, I thought I'd do the same for i^2 and got

n^2(n^2 + 1)/4

Either this is correct or I'm not sure where you're headed.

Well, the second part isn't correct at all ($\sum^{n}_{i=1} i^2 = \frac{n(n+1)(2n+1)}{6}$), but that wasn't really my point. Look at the left hand side of your proposed equality.

 

[Shackman]

Alright, so the left hand side is equal to ((n(n+1))/2)^2 or (n^4 +2n^3 +n^2)/4...

 

[Mystic998]

So prove that (n(n+1)/2)^2 is equal to the right hand side. Much easier to verify.

 

[Shackman]

That is a proof of the equality but it isn't inductive though. Inductive proofs always follow the form:
1) prove the base case
2) write the induction hypothesis for the case of n (assume its true basically)
3) use 2) to prove for n+1

 

[Hurkyl]

(1+2+...+n+(n+1))^2 != (1+2+...+n)^2 + (n+1)^2

As you say, that is not the correct formula for the square of two things. Fortunately, you know the right formula for the square of two things... (right?)

 

[Shackman]

You're talking about the binomial theorem right?