题目内容
数列{
}的前n项和为Sn，则
Sn=______．
| 1 |
| 4n2-1 |
| lim |
| n→∞ |
∵an=
=
=
(
-
)
∴Sn=a1+a2+…+an
=
[(1-
)+(
-
)+…+(
-
)]
=
(1-
)
∴
Sn=
(1-
)=
．
故答案为：
| 1 |
| 4n2-1 |
| 1 |
| (2n-1)•(2n+1) |
| 1 |
| 2 |
| 1 |
| 2n-1 |
| 1 |
| 2n+1 |
∴Sn=a1+a2+…+an
=
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 5 |
| 1 |
| 2n-1 |
| 1 |
| 2n+1 |
=
| 1 |
| 2 |
| 1 |
| 2n+1 |
∴
| lim |
| n→∞ |
| lim |
| n→∞ |
| 1 |
| 2 |
| 1 |
| 2n+1 |
| 1 |
| 2 |
故答案为：
| 1 |
| 2 |
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