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证明:(1)当n=1时,左= 1 3=右,等式成立. (2)假设当n=k时等式成立, 即 12 1•3+ 22 3•5+…+ k2 (2k−1)(2k+1)= k(k+1) 2(2k+1), 当n=k+1时,左边= 12 1•3+ 22 3•5+…+ k2 (2k−1)(2k+1)+ (k+1)2 (2k+1)(2k+3)= k(k+1) 2(2k+1)+ (k+1)2 (2k+1)(2k+3)= (k+1)(k+2) 2(2k+3). ∴当n=k+1时,等式也成立. 综合(1)(2),等式对所有正整数都成立.
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