<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] $${ 1 }^{ 2 }-{ 2 }^{ 2 }=-\left( 1+2 \right) \\ { 3 }^{ 2 }-{ 4 }^{ 2 }=-\left( 3+4 \right) \\ { 5 }^{ 2 }-{ 6 }^{ 2 }=-\left( 5+6 \right) \\ \vdots \\ { \left( 2n-1 \right)  }^{ 2 }-{ \left( 2n \right)  }^{ 2 }=-\left( 2n-1+2n \right) $$
Sum up, you get the answer that O.L. has written.
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