已知数列{an}的前n项和是Sn,且Sn=2an-n(n∈N*). (1)证明:数列{an+1}是等比数列,并求数列{an}的通项公式; (2)记bn=an+1anan+1,求数列{bn}的前n项和.

问题描述:

已知数列{an}的前n项和是Sn,且Sn=2an-n(n∈N*).
(1)证明:数列{an+1}是等比数列,并求数列{an}的通项公式;
(2)记bn=

an+1
anan+1
,求数列{bn}的前n项和.

(1)∵Sn=2an-n,∴n=1时,a1=2a1-1,解得a1=1.n≥2时,an=Sn-Sn-1=(2an-n)-(2an-1-n+1)=2an-2an-1-1,∴an=2an-1+1,∴an+1=2(an-1+1),∵a1+1=2,∴数列{an+1}是首项为2,公比为2的等比数列.∴an+1=2n,...