[再寄小读者之数学篇](2014-11-19 关于平方数的交叉和的两个代数等式)

For $ngeq 1$ to be an integer, $$ex (2n)^2-(2n+1)^2+cdots+(4n)^2 =-(4n+1)^2+cdots+(6n)^2, eex$$ $$ex (2n+1)^2-(2n+2)^2+cdots+(4n-1)^2 =-(4n)^2+(4n+1)^2-cdots+(6n-1)^2. eex$$ Ref. [Proof Without Words: Alternating Sums, The College Mathematics Journal].