1的二次方+2的二次方+3的二次方+...+n的二次方怎么算
人气:239 ℃ 时间:2020-02-04 07:58:26
解答
i^2 = i(i+1)-i= (1/3)[i(i+1)(i+2) - (i-1)i(i+1)] -(1/2)[i(i+1)- (i-1)i]1^2+2^2+.+n^2=∑(i:1->n) i^2=∑(i:1->n) {(1/3)[i(i+1)(i+2) - (i-1)i(i+1)] -(1/2)[i(i+1)- (i-1)i]}= (1/3)n(n+1)(n+2) -(1/2)n(n+1)=...
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