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 forgot a minus sign.
$\displaystyle P(k+1):1^2-2^2+3^2-.....+(-1)^{k-1}k^2+(-1)^k(k+1)^2\\=(-1)^{k-1}\dfrac{k.(k+1)}{2}+(-1)^k(k+1)^2\\=\dfrac{(k+1)}{2} [(-1)^{k-1}.k +2(-1)^k (k+1)]\\=\dfrac{(k+1)}{2} [-(-1)^{k}.k +2(-1)^k (k+1)]\\=(-1)^k\dfrac{(k+1)}{2}[-k+2k+2]\\=(-1)^k\dfrac{(k+1)}{2}[k+2]\\=(-1)^k\dfrac{(k+1)(k+2)}{2}$ 
(Proved)
