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??

For odd $n$, $(-1)^{n-1}=1$, so
$$1^2-2^2+3^2-4^2+\cdots+n^2\\
=1^2+(-2^2+3^2)+(-4^2+5^2)+\cdots+(-(n-1)^2+n^2)\\
=1+\color{blue}{(-2+3)}(2+3)+\color{blue}{(-4+5)}(4+5)+\cdots+\color{blue}{(-(n-1)+n)}((n-1)+n)\\
=\color{blue}{1\cdot}\left(1+2+3+4+5+\cdots+(n-1)+n\right)\\
=\frac {n(n+1)}2\\
=(-1)^{n-1}\frac {n(n+1)}2$$
For even $n$, $(-1)^{n-1}=-1$, so
$$1^2-2^2+3^2-4^2+\cdots+n^2\\
=(1^2-2^2)+(3^2-4^2)+(5^2-6^2)+\cdots+((n-1)^2-n^2)\\
=\color{orange}{(1-2)}(1+2)+\color{orange}{(3-4)}(3+4)+\color{orange}{(5-6)}(5+6)+\cdots+\color{orange}{((n-1)-n)}((n-1)+n)\\
=\color{orange}-(1+2+3+4+5+\cdots+(n-1)+n)\\
=-\frac {n(n+1)}2\\
=(-1)^{n-1}\frac {n(n+1)}2$$
Hence, for both odd and even $n$, 
$$1^2-2^2+3^2-4^2+\cdots+n^2=(-1)^{n-1}\frac {n(n+1)}2$$
