Prove 0² + 1² + 2² + ... + n² = n(n+1)(2n+1)/6 by induction

[mcraze123]

Homework Statement

prove:
0^2 + 1^2 + 2^2 + ... + n^2 = n(n+1)(2n+1)/6

Homework Equations

The Attempt at a Solution

I'm confused on how to prove this by induction. I'm not exactly sure what the goal of the rearrangement is after substituting (n+1). Any help is much appreciated!

base case: n = 0
0^2 = 0(0+1)(2*0+1)/6

induction step:
(0^2+1^2+2^2+...+n^2) + (n+1)^2 = (n+1)((n+1)+1)(2(n+1)+1)/6

n(n+1)(2n+1)/6 + (n+1)^2

(n+1)[n(2n+1)/6 + (n+1)]
...here is where I'm lost, I'm not sure what I'm trying to manipulate it to look like...

Thanks!

 

[rock.freak667]

(n+1)[n(2n+1)/6 + (n+1)]
...here is where I'm lost, I'm not sure what I'm trying to manipulate it to look like...

Factor out 1/6 and then simplify what is left in the brackets.

 

[vela]

induction step:
(0^2+1^2+2^2+...+n^2) + (n+1)^2 = (n+1)((n+1)+1)(2(n+1)+1)/6

n(n+1)(2n+1)/6 + (n+1)^2

(n+1)[n(2n+1)/6 + (n+1)]
...here is where I'm lost, I'm not sure what I'm trying to manipulate it to look like...

You're trying to show the first equation is true. To do this, you start with one side, as you have done, and manipulate it until it looks like the other side.