问题
解答题
数列{an}的前n项和为Sn,Sn=2an-2.
(Ⅰ)求数列{an}的通项公式;
(Ⅱ)设bn=log34an,求数列{bn}的前n项和Tn.
答案
(Ⅰ)当n=1时,S1=2a1-2,∴a1=2,
当n≥2时,Sn=2an-2,Sn-1=2an-1-2
∴Sn-Sn-1=2an-2an-1=an
∴an=2an-1(n≥2),
∴数列{an}是首项为2,公比为2的等比数列
∴an=2n.
(Ⅱ)bn=log34an=log34•2n=log32n+2=(n+2)log32,
∴Tn=
log32.n(n+5) 2