1^2+2^2+3^2+-----+n^2=n(n+1)(2n+1)/6的推导?

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1^2+2^2+3^2+-----+n^2=n(n+1)(2n+1)/6的推导?

1^2+2^2+3^2+-----+n^2=n(n+1)(2n+1)/6的推导?
1^2+2^2+3^2+-----+n^2=n(n+1)(2n+1)/6的推导?

1^2+2^2+3^2+-----+n^2=n(n+1)(2n+1)/6的推导?
用数学归纳法
1.当n=1时,1^2=1=1*2*3/6
2.假设当n=k时,1^2+2^2+……+k^2=k(k+1)(2k+1)/6
则当n=k+1时,
1^2+2^2+……+k^2+(k+1)^2
=k(k+1)(2k+1)/6+(k+1)^2
=(k+1)(k+2)(2k+3)/6
假设成立
所以1^2+2^2+3^2+-----+n^2=n(n+1)(2n+1)/6

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