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??

$$\sum_{r=1}^{2n}(-1)^{r-1}r^2=\sum_{r=1}^n\{(2r-1)^2-(2r)^2\}=\sum_{r=1}^n(1-4r)$$
$$=\dfrac n2(1-4+1-4n)=\cdots=(-1)^{2n-1}\dfrac{2n(2n+1)}2$$
Now $$\sum_{r=1}^{2n+1}(-1)^{r-1}r^2=(2n+1)^2+\sum_{r=1}^{2n}(-1)^{r-1}r^2=?$$
