设数列{an}的前n项和sn,已知a1=2,sn=an+2^n-2,设bn=log2an,求数列{bn}的前n项和Tn

题目

设数列{an}的前n项和sn,已知a1=2,sn=an+2^n-2,设bn=log2an,求数列{bn}的前n项和Tn

答案

sn=an+2^n-2,sn-1=an-1+2^(n-1)-2,an=sn-sn-1=an+2^n-2-[an-1+2^(n-1)-2]=an+2^n-an-1-2^(n-1)an-1=2^n-2^(n-1)=2*2^(n-1)-2^(n-1)=2^(n-1)即an=2^nbn=log2anb1=log2(2^1)=1b2=log2(2^2)=2b3=log2(2^3)=3.bn=log2(2^...