已知数列{an}的前n项和为Sn,且Sn=2n^2+n,n∈N*,数列{bn}满足an=4log2(bn),n∈N*

题目

已知数列{an}的前n项和为Sn,且Sn=2n^2+n,n∈N*,数列{bn}满足an=4log2(bn),n∈N*
(1)求:数列{an},{bn}的通项公式；
(2)求:数列{an乘以bn}的前n项和Tn.

答案

(1)Sn=2n^2+n (1)n=1, =>a1=3S(n-1)=2(n-1)^2+(n-1) (2)(1)-(2)an = 4n-1an =4logbnbn = 2^(an/4) = 2^[(4n-1)/4](2)an.bn=(4n-1) 2^[(4n-1)/4] =2^(9/2). (n.2^n) - 2^[(4n-1...