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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
人气:419 ℃ 时间:2019-10-17 05:35:47
解答
(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
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