题目内容
若数列{an}的前n项和为Sn,且满足Sn=
an+1(n≥1),则an=______,
(a1+a2+a3+…+an)的值是______.
| 1 |
| 4 |
| lim |
| n→∞ |
由于数列{an}的前n项和为Sn,且满足Sn=
an+1(n≥1)①,令n=1可得a1=
.
当n≥2时,Sn-1=
an-1+1 ②,用①减去②,化简可得an=-
an-1,故数列为等比数列,公比为-
,∴an=
(-
)n-1.
∴Sn=
=1-(-
)n,∴
(a1+a2+a3+…+an)=
Sn=
[1-(-
)n]=1,
故答案为
(-
)n-1、1.
| 1 |
| 4 |
| 4 |
| 3 |
当n≥2时,Sn-1=
| 1 |
| 4 |
| 1 |
| 3 |
| 1 |
| 3 |
| 4 |
| 3 |
| 1 |
| 3 |
∴Sn=
| ||||
1+
|
| 1 |
| 3 |
| lim |
| n→∞ |
| lim |
| n→∞ |
| lim |
| n→∞ |
| 1 |
| 3 |
故答案为
| 4 |
| 3 |
| 1 |
| 3 |
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