数列{an}首项a1=1,前n项和Sn与an之间满足an=2Sn22Sn−1(n≥2) (1)求证:数列{1/Sn}是等差数列 (2)求数列{an}的通项公式.
问题描述:
数列{an}首项a1=1,前n项和Sn与an之间满足an=
(n≥2)2Sn2
2Sn−1
(1)求证:数列{
}是等差数列 1 Sn
(2)求数列{an}的通项公式.
答
(1)∵当n≥2时,an=Sn−Sn−1=
,2
S
2n
2Sn−1
整理得:Sn-1-Sn=2Sn⋅Sn-1,
由题意知Sn≠0,
∴
−1 Sn
=2,1 Sn−1
即{
}是以1 Sn
=1 S1
=1为首项,公差d=2的等差数列.1 a1
(2)∵{
}是以1 Sn
=1 S1
=1为首项,公差d=2的等差数列.1 a1
∴
=1+2(n−1)=2n−1,1 Sn
∴Sn=
,n∈N•.,1 2n−1
当n≥2时,an=Sn−Sn−1=
−1 2n−1
=−1 2(n−1)−1
,2 (2n−1)(2n−3)
当n=1时,a1=S1=1不满足an,
∴an=
.
1,n=1 −
,n≥22 (2n−1)(2n−3)