## Inductive Proof on Well Known Sum

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..
 PhysOrg.com science news on PhysOrg.com >> 'Whodunnit' of Irish potato famine solved>> The mammoth's lament: Study shows how cosmic impact sparked devastating climate change>> Curiosity Mars rover drills second rock target
 What is $$\sum^{n}_{i=1} i$$?
 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.

## Inductive Proof on Well Known Sum

 Quote by 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 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.
 Alright, so the left hand side is equal to ((n(n+1))/2)^2 or (n^4 +2n^3 +n^2)/4...
 So prove that (n(n+1)/2)^2 is equal to the right hand side. Much easier to verify.
 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

Recognitions:
Gold Member