问题描述: bn=(2n+1)乘2的(2n+1)次方的前n项和 1个回答 分类:数学 2014-10-28 问题解答: 我来补答 letS = 1.2^1+2.2^2+...+n.2^n (1)2S = 1.2^2+2.2^3+...+n.2^(n+1) (2)(2)-(1)S = n.2^(n+1) - (2^1+2^2+...+2^n)=n.2^(n+1) - (2^n -1 )4S =4n.2^(n+1) - 4(2^n -1 )bn = (2n+1).2^(n+1)= 4(n.2^n) + 2^(n+1)Tn = b1+b2+...+bn= 4S + 4(2^n-1)=4n.2^(n+1)=8n.2^n 展开全文阅读