# Inductive Proof on Well Known Sum

1. Feb 25, 2008

### Shackman

1. The problem statement, all variables and given/known data
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..

2. Feb 25, 2008

### Mystic998

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

3. Feb 25, 2008

### 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.

Last edited: Feb 25, 2008
4. Feb 25, 2008

### Mystic998

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.

5. Feb 25, 2008

### Shackman

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

6. Feb 25, 2008

### Mystic998

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

7. Feb 25, 2008

### 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

8. Feb 25, 2008

### Hurkyl

Staff Emeritus
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?)

9. Feb 25, 2008

### Shackman

You're talking about the binomial theorem right?