How to deal with $(-1)^{k-1}$ It's a problem on mathematical induction.   

$$1^2-2^2+3^2-.....+(-1)^{n-1}n^2=(-1)^{n-1}\frac{n.(n+1)}{2}$$  

I have proved it for values of $n=1,2$. 
Now I assume for $n=k$  
$$P(k):1^2-2^2+3^2-.....+(-1)^{k-1}k^2=(-1)^{k-1}\frac{k.(k+1)}{2}$$.  
$$P(k+1):1^2-2^2+3^2-.....+(-1)^{k-1}k^2+(k+1)^2=(-1)^{k-1}\frac{k.(k+1)}{2}+(k+1)^2\\=\frac{(k+1)}{2} [(-1)^{k-1}.k+2k+2]$$  
I need suggestion to deal with the $(-1)^{k-1}$ so that I can prove the whole. Any help is appreciated.

you have to prove that $$(-1)^k\frac{k(k+1)}{2}+(-1)^k(k+1)^2=(-1)^k\frac{(k+1)(k+2)}{2}$$ or
$$(-1)^{k-1}\frac{k(k+1)}{2}=(-1)^k\frac{(k+1)(k+2)}{2}-(-1)^k(k+1)^2$$
the right-hand side is given by
$$(-1)^k(k+1)\left(\frac{(k+2)}{2}-k-1\right)$$ and this is equal to
$$(-1)^k(k+1)\left(\frac{k+2-2k-2}{2}\right)$$ Can you finish?
