数列{an}的前n项和为Sn,且Sn=13(an−1)(1)求a1,a2及a3;(2)证明:数列{an}是等比数列,并求an.

问题描述:

数列{an}的前n项和为Sn,且Sn

1
3
(an−1)
(1)求a1,a2及a3;(2)证明:数列{an}是等比数列,并求an

(1)当n=1时,a1S1

1
3
(a1−1),得a1=−
1
2

当n=2时,S2a1+a2
1
3
(a2−1)
,得a2
1
4
,同理可得a3=−
1
8

(2)当n≥2时,anSnSn−1
1
3
(an−1)−
1
3
(an−1−1)=
1
3
an
1
3
an−1

2
3
an=−
1
3
an−1

所以
an
an−1
=−
1
2

a1=−
1
2
,∴an=(−
1
2
)n

故数列{an}是等比数列,an=(−
1
2
)n

答案解析:(1)在Sn
1
3
(an−1)
中,分别令n=1,2,3,能够求出a1,a2及a3的值.
(2)当n≥2时,anSnSn−1
1
3
(an−1)−
1
3
(an−1−1)=
1
3
an
1
3
an−1
,所以
an
an−1
=−
1
2
.由此能求出an
考试点:数列的应用.

知识点:本题考查数列的性质和应用,解题时要认真审题,仔细求解.