题目内容
数列{an}中,a3=1,a1+a2+…+an=an+1(n=1,2,3…).
(Ⅰ)求a1,a2;
(Ⅱ)求数列{an}的前n项和Sn;
(Ⅰ)求a1,a2;
(Ⅱ)求数列{an}的前n项和Sn;
(Ⅰ)∵a1=a2,a1+a2=a3,
∴2a1=a3=1,
∴a1=
,a2=
.
(Ⅱ)∵Sn=an+1=Sn+1-Sn,∴2Sn=Sn+1,
=2,
∴{Sn}是首项为S1=a1=
,公比为2的等比数列.
∴Sn=
•2n-1=2n-2.
∴2a1=a3=1,
∴a1=
| 1 |
| 2 |
| 1 |
| 2 |
(Ⅱ)∵Sn=an+1=Sn+1-Sn,∴2Sn=Sn+1,
| Sn+1 |
| Sn |
∴{Sn}是首项为S1=a1=
| 1 |
| 2 |
∴Sn=
| 1 |
| 2 |
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| 1 |
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B、
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