<s>[INST] Find the sum of the series $1^2-2^2+3^2-4^2+...-(2n)^2$ Find the sum of the series $$1^2-2^2+3^2-4^2+...-(2n)^2$$
I tried rewriting it as $$\sum_{r=1}^{2n}-1^{n+1}(r^2)$$ but it didn't help.
Also, looked at re-arranging as $$1^2+3^2+5^2+7^2+...+(2n-1)^2$$ and $$-2^2-4-6^2-8^2-...-(2n)^2$$
Still couldn't get to the given answer of $-n(2n+1)$
 [/INST] Note that
$$(2n-1)^2-(2n)^2=1-4n,$$
and therefore your sum is equal to
$$\underbrace{(1+1+\ldots+1)}_{n\;\mathrm{times}}-4\left(1+2+\ldots+n\right)=n-4\cdot\frac{n(n+1)}{2}=-n(2n+1).$$
The only thing one needs to know is the arithmetic progression sum.
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