Prove that $$1^2-2^2+3^2...+(-1)^{n-1}n^2=(-1)^{n-1}\frac{n(n+1)}{2}$$ whenever $n$ is a positive integer.
I used $2$ as my base case and it worked. Then I plugged in $k$ for $n$. Now I can't figure out how to do the $k+1$ step. I would greatly appreciate some help.
It's generally a better idea to assume it's true for $n$, then plug in $n+1$ and see what you can do, rather than writing it down for $n$ and then trying to somehow turn it into the $n+1$ case.
So assume it's true for $n$ and write down the sum for $n+1$:
$$\sum_{k=1}^{n+1}(-1)^{k-1}k^2$$
The nice things about sums and induction is that the case for $n$ easily breaks away (another reason why writing down the $n+1$ case is usually a better approach):
$$=\left(\sum_{k=1}^{n}(-1)^{k-1}k^2\right)+(-1)^n(n+1)^2=(-1)^{n-1}\frac{n(n+1)}{2}+(-1)^n(n+1)^2$$
Now, the expression you want will have a factor of $(-1)^n$, so factor one of those out and see what happens:
$$=(-1)^{n}\left(-\frac{n(n+1)}{2}+(n+1)^2\right)$$
Now just follow your nose...
$$n=2m(+1)\iff S=\sum_1^nk^2-2\cdot2^2\sum_1^{m}k^2=\frac{n(n+1)}2-4m(m+1).$$
