已知数列{an}的首项a1=3,通项an与前n项和Sn之间满足2an=SnSn-1(n≥2).(1)求证{1/Sn}是等差数列,并求公差;(2)求数列{an}的通项公式.

问题描述:

已知数列{an}的首项a1=3,通项an与前n项和Sn之间满足2an=SnSn-1(n≥2).
(1)求证{

1
Sn
}是等差数列,并求公差;
(2)求数列{an}的通项公式.

(1)∵2an=SnSn-1(n≥2)∴2(Sn-Sn-1)=SnSn-1
两边同时除以SnSn-1,得2(

1
Sn-1
-
1
Sn
)=1
1
Sn
-
1
Sn-1
=-
1
2

{
1
Sn
}
是等差数列,公差d=-
1
2

(2)∵
1
S1
=
1
a1
=
1
3

1
Sn
=
1
3
+(n-1)×(-
1
2
)=-
1
2
n+
5
6
=
5-3n
6

Sn=
6
5-3n

当n≥2时,an=
1
2
SnSn-1=
1
2
×
6
5-3n
×
6
8-3n
=
18
(5-3n)(8-3n)

an=
3 ,n=1
18
(8-3n)(5-3n)
,n≥2