求数列{(2n-1)*1/4的n次方}的前n项和Sn
人气:387 ℃ 时间:2020-03-26 15:25:00
解答
an = (2n-1)(1/4)^n= n(1/4)^(n-1) - (1/4)^nSn =a1+a2+..+an= [summation(i:1->n){i(1/4)^(i-1)} ] - (1/3)(1- (1/4)^n)consider[x^(n+1)-1]/(x-1) = 1+x+x^2+...+x^n[(x^(n+1)-1)/(x-1)]' = 1+2x+..+nx^(n-1)1+2x+...
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