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.
 A: You are making a mistake, in that you are assuming that $(-1)^k=1$. 
You have 
\begin{align}
(-1)^{k-1}\frac{k.(k+1)}{2}+(-1)^k(k+1)^2&=(-1)^{k-1}\frac{(k+1)}{2} [k-(2k+2)]\\ \ \\
&=(-1)^{k-1}\frac{(k+1)}{2} [-(k+2)]\\ \ \\
&=(-1)^{k}\frac{(k+1)(k+2)}{2} \\ \ \\
\end{align}
A: L.H.S. of $P(k+1)$ should be $1^2-2^2+3^2-\cdots+(-1)^{k-1}k^2+(-1)^k(k+1)^2$
which is equal to 
$\displaystyle \frac{(-1)^{k-1}k(k+1)}{2}+(-1)^k(k+1)^2=\frac{(-1)^{k}[-k(k+1)+2(k+1)^2]}{2}=\frac{(-1)^{k}(k+1)(k+2)}{2}$
A: 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}
A: 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?
A: 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)
