Sum of terms $1^2-2^2+3^2-...$ Prove that: $$1^2-2^2+3^2-4^2+...+(-1)^{n-1}n^2=\frac{(-1)^{n-1}n(n+1)}{2}$$
I proved it by induction but is there any other way to solve it?
If it was not a proof but rather a question like find the term,how to solve it?
I realized that alternate terms were under same sign but can't understand whether to take $\frac{n}{2}$ odd and even terms[if n is even] or $\frac{n+1}{2}$ odd terms and $\frac{n-1}{2}$ even terms[if n is odd]. 
I thought of this $$1^2-2^2+3^2-4^2+5^2=1^2+{(1+2)}^2+{(1+4)}^2-2^2-4^2$$
$$=1^2+1^2+1^2+2(0+2+4)+2^2+4^2-2^2-4^2$$
But then how to generalize??

Note:  $(n+1)^2 = n^2 + 2n +1$.
So $(2n-1)^2 - (2n)^2 = 4n^2 -4n +1 - 4n^2 = -4n+1$.
So $\sum_{k=1}^{n=2m} (-1)^{k-1}k^2 = \sum_{j=1}^m [(2j-1)^2 - (2j)^2]=\sum_{j=1}^m [-4j +1]= -4(\sum j) + m = -4*\frac {m(m+1)}2 + m = -2m(m+1) +m=m(-2(m+1) + 1)=m(-2m-1)=-m(2m+1)= -\frac{n(n+1)}{2}$
..... if $n$ is even.
If $n=2m+1$ is odd then $\sum_{k=1}^{n=2m+1}(-1)^{k-1}k^2 = [\sum_{k=1}^{2m=2n-1}(-1)^{k-1}k^2] + n^2$
$= -\frac {(n-1)((n-1) + 1)}2 + n^2= n^2 - \frac {n(n-1)}2 = \frac {2n^2 -n(n-1)}2 = \frac {n(2n - (n-1)}2 = \frac {n(n+1)}2$
So either way ...
