<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] Hint 1: We have $$\sum_{k=1}^nk^2=\frac{1}{6}n(n+1)(2n+1).$$
Hint 2: Note that 
$$1^2-2^2+3^2-4^2+\cdots-(2n)^2=1^2+2^2+\cdots+(2n)^2-2(2^2+4^2+\cdots+(2n)^2)$$
$$
=1^2+2^2+\cdots+n^2-2^3(1^2+2^2+\cdots+n^2)
=\sum_{k=1}^{2n}k^2-8\sum_{k=1}^{n}k^2.$$
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