Question

当n∈N*时, 22+42+…+(2n)2＝n(n+1)(2n+1)

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Answer 1

答案：T
解析：

证明:(1)当n＝1时, ∵左边＝22＝4, 右边＝×1×2×3＝4,          ∴ 等式成立      (2)假设n＝k(k∈N)时等式成立.         即 22+42+…+(2k)2＝k(k+1)(2k+1)         ∵   22+42+…+(2k)2+［2(k+1)］2            ＝k(k+1)(2k+1)+4(k+1)2            ＝(k+1)(2k2+k+6k+6)            ＝(k+1)(2k2+7k+6)            ＝(k+1)(k+2)(2k+3)            ＝(k+1)［(k+1)+1］［2(k+1)+1］         ∴ 当n＝k+1时等式仍然成立.根据(1),(2),          ∴ 对于一切n∈N等式成立.

提示：

k(k+1)(2k+1)+4(k+1)2＝(k+1)［k(2k+1)+6(k+1)］