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=
(an−1)1 3
(1)求a1,a2及a3;(2)证明:数列{an}是等比数列,并求an.
答
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.
(1)当n=1时,a1=S1=
(a1−1),得a1=−1 3
;1 2
当n=2时,S2=a1+a2=
(a2−1),得a2=1 3
,同理可得a3=−1 4
.1 8
(2)当n≥2时,an=Sn−Sn−1=
(an−1)−1 3
(an−1−1)=1 3
an−1 3
an−1,1 3
∴
an=−2 3
an−1,1 3
所以
=−an an−1
.1 2
∵a1=−
,∴an=(−1 2
)n.1 2
故数列{an}是等比数列,an=(−
)n.1 2
答案解析:(1)在Sn=
(an−1)中,分别令n=1,2,3,能够求出a1,a2及a3的值.1 3
(2)当n≥2时,an=Sn−Sn−1=
(an−1)−1 3
(an−1−1)=1 3
an−1 3
an−1,所以1 3
=−an an−1
.由此能求出an.1 2
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.