Proof by induction that $\sum_{k=1}^n (-1)^{k-1}k^2 = (-1)^{n-1}\frac{(n)(n+1)}{2}$ Through Induction I tried to prove that: 
$$\sum_{k=1}^n (-1)^{k-1}k^2 = (-1)^{n-1}\frac{(n)(n+1)}{2}$$
I first let $n=1$, so that on the left hand side and the right hand side you get 1.
Then I tried to prove that this also works when $$n=n+1$$
So the equation then becomes:
$$\sum_{k=1}^{n+1} (-1)^{k-1}k^2 = \sum_{k=1}^n (-1)^{k-1}k^2 +(-1)^{n-1}\frac{(n)(n+1)}{2}$$
As we know that $\sum_{k=1}^n (-1)^{k-1}k^2 = (-1)^{n-1}\frac{(n)(n+1)}{2}$, we can replace the first term after the $=$ with this. The equation becomes:
$$=(-1)^{n-1}\frac{(n)(n+1)}{2} + (-1)^{n-1}\frac{(n)(n+1)}{2}$$
The I subsitute $n+1$ for $n$, so that:
$$=(-1)^{n+1-1}\frac{(n+1)(n+1+1)}{2} + (-1)^{n+1-1}\frac{(n+1)(n+1+1)}{2}$$
$$=(-1)^{n}\frac{(n+1)(n+2)}{2} + (-1)^{n}\frac{(n+1)(n+2)}{2}$$
$$=(-1)^{n}[\frac{(n+1)(n+2)}{2}+\frac{(n+1)(n+2)}{2}]$$
$$=(-1)^{n}[\frac{(2n^2+6n+4)}{2}]$$
$$=(-1)^{n}[\frac{(2n^2+2n+4n+4)}{2}]$$
$$=(-1)^{n}[\frac{((n+1)(2n+4))}{2}]$$
But how do I proceed from here?

i think it must be
$$\sum_{k=1}^{n+1}(-1)^{k-1}k^2=\sum_{k=1}^n(-1)^{k-1}k^2+(-1)^n(n+1)^2$$
and we get $$(-1)^{n-1} \frac {n(n+1)}{2}+(-1)^n(n+1)^2=(-1)^n\frac{(n+1)(n+2)}{2}$$
