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This question already has an answer here:

I want to find the sum of a series for which the $n^{th}$ term is given by $T_n=n(n+1)(-1)^{n+1}$ Basically if we find the sum of $n$ terms the series turns out to be like this: $1 \times 2 -2 \times 3 +3 \times 4 -4 \times 5 +5 \times 6\ldots$

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marked as duplicate by Siong Thye Goh, Lord Shark the Unknown, JMP, José Carlos Santos, Claude Leibovici Aug 21 '17 at 7:12

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If $n=2m$ is even, the sum is $$2(1-3)+4(3-5)+6(5-7)+\cdots+2m((2m-1)-(2m+1)) =2(-2)+4(-2)+\cdots+2m(-2) =-4(1+2+\cdots+m)$$ which I hope is a familiar sum. I'll leave the $n$ odd case to you.

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