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

2Sn2
2Sn−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.

(1)∵当n≥2时,anSnSn−1

2
S 2n
2Sn−1

整理得:Sn-1-Sn=2Sn⋅Sn-1
由题意知Sn≠0,
1
Sn
1
Sn−1
=2

{
1
Sn
}
是以
1
S1
1
a1
=1
为首项,公差d=2的等差数列.
(2)∵{
1
Sn
}
是以
1
S1
1
a1
=1
为首项,公差d=2的等差数列.
1
Sn
=1+2(n−1)=2n−1

Sn
1
2n−1
,n∈N
.,
当n≥2时,anSnSn−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
2
(2n−1)(2n−3)
,n≥2