Okay, I'm not gonna detail this summation proof, but this is what the inductive step comes down to:

I have to prove that $\displaystyle (k^2-1) + (2(k+1) +1)$ is the same as $\displaystyle (k+1)^2 -1$

$\displaystyle (k^2-1) + (2(k+1) +1)$

$\displaystyle =k^2 -1 + 2k + 2 + 1$

$\displaystyle = k^2 + 2k + 2$

Since this can't factor, I took made it look like this:

$\displaystyle =[k^2 + 2k + 1] +1$

$\displaystyle = [(k+1)(k+1)]+1$

$\displaystyle = (k+1)^2 + 1$

So I can't seem to prove that it's $\displaystyle (k+1)^2-1$, but the +1 algebra works...

In order for this proof to work, it has to be -1, anyone see anything wrong with what I did?

Thanks.