设数列an的前n项和为Sn 已知a1+2a2+3a3+……+nan=(n-1)Sn+2n

题目

设数列an的前n项和为Sn 已知a1+2a2+3a3+……+nan=(n-1)Sn+2n
抽去{an}中a1,a4,a7.a(3n-2),.余下的顺序不变组成{bn},若{bn}前n项和为Tn,求证12/5＜T(n+1）/Tn≤11/3

答案

由a1+2a2+3a3+……+nan=(n-1)Sn+2n可知：n=1时：a1=(1-1)s1+2,解得：a1=2；n=2时：a1+2a2=(2-1)s2+4,即2+2a2=(2+a2)+4,解得：a2=4.由题意,有：a1+2a2+3a3+……+nan=(n-1)Sn+2na1+2a2+3a3+……+nan+(n+1)a(n+1)=[(n+1...