题目内容
已知数列{an} 的前n项和Sn ,且Sn=
(an-1)(n∈N*).
(1)求a1,a2,a3;
(2)求证:数列{an} 是等比数列.
| 1 |
| 3 |
(1)求a1,a2,a3;
(2)求证:数列{an} 是等比数列.
(1)∵Sn=
(an-1),
∴S1=
(a1-1),∴a1=-
;
∵S2=
(a2-1),∴a2=
;
∵S3=
(a3-1),∴a3=-
;
(2)证明:∵Sn=
(an-1),
∴Sn-1=
(an-1-1),
两式相减:an=
(an-an-1),
∴当n≥2时,
=-
,
∴数列{an}是等比数列.
| 1 |
| 3 |
∴S1=
| 1 |
| 3 |
| 1 |
| 2 |
∵S2=
| 1 |
| 3 |
| 1 |
| 4 |
∵S3=
| 1 |
| 3 |
| 1 |
| 8 |
(2)证明:∵Sn=
| 1 |
| 3 |
∴Sn-1=
| 1 |
| 3 |
两式相减:an=
| 1 |
| 3 |
∴当n≥2时,
| an |
| an-1 |
| 1 |
| 2 |
∴数列{an}是等比数列.
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