For $n\geq 1$ to be an integer, $$\bex (2n)^2-(2n+1)^2+\cdots+(4n)^2 =-(4n+1)^2+\cdots+(6n)^2, \eex$$ $$\bex (2n+1)^2-(2n+2)^2+\cdots+(4n-1)^2 =-(4n)^2+(4n+1)^2-\cdots+(6n-1)^2. \eex$$ Ref. [Proof Without Words: Alternating Sums, The College Mathematics Journal].
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原文：http://www.cnblogs.com/zhangzujin/p/4107690.html