Question

當(dāng)n∈N*時(shí), 22+42+…+(2n)2＝n(n+1)(2n+1)

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Answer 1

答案：T
解析：

證明:(1)當(dāng)n＝1時(shí), ∵左邊＝22＝4, 右邊＝×1×2×3＝4,          ∴ 等式成立      (2)假設(shè)n＝k(k∈N)時(shí)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］         ∴ 當(dāng)n＝k+1時(shí)等式仍然成立.根據(jù)(1),(2),          ∴ 對(duì)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)］