Prove by induction: $\sum\limits_{k=1}^n (-1)^{k+1}k^2 = \frac{(-1)^{n+1}n(n+1)}{2}$ The whole problem has been translated from German, so apologies if I made any mistakes. Thank you for taking the time to help!
So this is a problem from my math book 
Prove that $\sum\limits_{k=1}^n (-1)^{k+1}k^2 = \frac{(-1)^{n+1}n(n+1)}{2}$ for all $n\  \varepsilon \ N$
and I have come as far as this:


*

*let $n_0 = 1$. Then $\sum\limits_{k=1}^n (-1)^{k+1}k^2 = (-1)^{1+1}1^2 = 1 = \frac{(-1)^{1+1}1(1+1)}{2} = 1 $

*For one $n\  \varepsilon \ N$  $\sum\limits_{k=1}^n (-1)^{k+1}k^2 = \frac{(-1)^{n+1}n(n+1)}{2}$ is true.

*Now we have to show that  $\sum\limits_{k=1}^{(n+1)} (-1)^{k+1}k^2 = \frac{(-1)^{(n+1)+1}(n+1)((n+1)+1)}{2} = \sum\limits_{k=1}^{(n+1)} (-1)^{k+1}k^2 = \frac{(-1)^{(n+2)}(n+1)((n+2)}{2}$ for all $n\  \varepsilon \ N$ 
From here, I did this: $\sum\limits_{k=1}^{(n+1)} (-1)^{k+1}k^2 = \sum\limits_{k=1}^{(n)} (-1)^{k+1}k^2 + (-1)^{(n+2)}(n+1)^2 = \frac{(-1)^{n+1}n(n+1)}{2} + (-1)^{(n+2)}(n+1)^2 = \frac{(-1)^{n+1}n(n+1)}{2} + \frac{2((-1)^{(n+2)}(n+1)^2)}2$ 
at which point I got stuck.
A look at the solutions at the back of the book told me that the next step is to factor (n+1) and $(-1)^{(n+2)}$, which resulted into this: 
$\frac{(-1)^{n+2}(n+1)((-1)n+2(n+1))}{2} = \frac{(-1)^{n+2}(n+1)(-n+2n+2)}{2} = \frac{(-1)^{n+2}(n+1)(n+2)}{2}$ 
Which makes sense, but here is my question: How on earth do I get from $\frac{(-1)^{n+1}n(n+1)}{2} + \frac{2((-1)^{(n+2)}(n+1)^2)}2$ to $\frac{(-1)^{n+2}(n+1)((-1)n+2(n+1))}{2}$? I have been trying for quite a while and I just can't seem to figure it out. Any help is greatly appreciated!
 A: From here
$$= \frac{(-1)^{n+1}n(n+1)}{2} + \frac{2((-1)^{(n+2)}(n+1)^2)}2
=(-1)^{(n+2)}\left(\frac{-n(n+1)}{2} +\frac{2(n+1)^2)}2\right)
=(-1)^{(n+2)}\left(\frac{(n+1)(n+2)}{2}\right)$$
A: Because if $$1^2-2^2+...+(-1)^n(n-1)^2=\frac{(-1)^nn(n-1)}{2}$$ then
$$1^2-2^2+...+(-1)^n(n-1)^2+(-1)^{n+1}n^2=$$
$$=(-1)^{n+1}\left(n^2-\frac{n(n-1)}{2}\right)=(-1)^{n+1}\frac{n(n+1)}{2}.$$
The base is obvious.
A: This is obviously true for $n=1$. Assume it's true for any positive integer $n$
$$\sum\limits_{k=1}^n (-1)^{k+1}k^2 = \frac{(-1)^{n+1}n(n+1)}{2}$$
we must show that:
$$\sum\limits_{k=1}^{n+1} (-1)^{k+1}k^2 = \frac{(-1)^{n+2}(n+1)(n+2)}{2}$$
from the other side we have:
$$\sum\limits_{k=1}^{n+1} (-1)^{k+1}k^2 =\sum\limits_{k=1}^n (-1)^{k+1}k^2+(-1)^{n+2}(n+1)^2 $$
by substitution from our induction assumption we obtain:
$$\sum\limits_{k=1}^{n+1} (-1)^{k+1}k^2=\frac{(-1)^{n+1}n(n+1)}{2}+(-1)^{n+2}(n+1)^2=\frac{(-1)^{n+2}(-n^2-n+2n^2+4n+2)}{2}=\frac{(-1)^{n+2}(n^2+3n+2)}{2}=\frac{(-1)^{n+2}(n+1)(n+2)}{2}$$
which is what we wnated to show.
