怎样推理:1^2+2^2+3^2+……+n^2=n(n+1)(2n+1)/6

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怎样推理:1^2+2^2+3^2+……+n^2=n(n+1)(2n+1)/6怎样推理:1^2+2^2+3^2+……+n^2=n(n+1)(2n+1)/6怎样推理:1^2+2^2+3^2+……+n^2=

怎样推理:1^2+2^2+3^2+……+n^2=n(n+1)(2n+1)/6
怎样推理:1^2+2^2+3^2+……+n^2=n(n+1)(2n+1)/6

怎样推理:1^2+2^2+3^2+……+n^2=n(n+1)(2n+1)/6
数学归纳法
首先,对n=1成立
其次,1^2+2^2+3^2+……+n^2+(n+1)^2
=n(n+1)(2n+1)/6+(n+1)^2
=[6(n+1)^2+n(n+1)(2n+1)]/6
=[(2n^2+n+6n+6)(n+1)]/6
=[(n+1)(n+2)(n+3)]/6
所以成立




:)