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.

If$$1^2-2^2+\cdots+(-1)^{k-1}k^2=(-1)^{k-1}\frac{k(k+1)}2,$$then\begin{align}1^2-2^2+\cdots+(-1)^k(k+1)^2&=(-1)^{k-1}\frac{k(k+1)}2+(-1)^k(k+1)^2\\&=(-1)^k\left(-\frac{k(k+1)}2+(k+1)^2\right)\\&=(-1)^k(k+1)\left(k+1-\frac k2\right)\\&=(-1)^k(k+1)\frac{k+2}2\\&=(-1)^k\frac{(k+1)(k+2)}2.\end{align}
