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1*4,2*7,3*30,……,n(3n+1)的前n项和
人气:470 ℃ 时间:2020-06-11 11:22:49
解答
∑n(3n+1)=∑(3n^2+n)=∑3n^2+∑n=3∑n^2+n(n+1)/2
=3[n(n+1)(2n+1)/6]+n(n+1)/2
=n(n+1)(2n+1)/2+n(n+1)/2
=n(n+1)(2n+2)/2
=n(n+1)(n+1)
=n(n+1)^2
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