题目内容
已知数列{an}满足an=
，则其前n项和为Sn=
．
| 1 |
| 4n2-1 |
| n |
| 2n+1 |
| n |
| 2n+1 |
分析：由an=
=
=
（
-
），利用裂项求和即可求解
| 1 |
| 4n2-1 |
| 1 |
| (2n-1)(2n+1) |
| 1 |
| 2 |
| 1 |
| 2n-1 |
| 1 |
| 2n+1 |
解答：解：∵an=
=
=
（
-
）
∴Sn=
(1-
+
-
+…+
-
)
=
(1-
)
=
故答案为：
| 1 |
| 4n2-1 |
| 1 |
| (2n-1)(2n+1) |
| 1 |
| 2 |
| 1 |
| 2n-1 |
| 1 |
| 2n+1 |
∴Sn=
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 5 |
| 1 |
| 2n-1 |
| 1 |
| 2n+1 |
=
| 1 |
| 2 |
| 1 |
| 2n+1 |
=
| n |
| 2n+1 |
故答案为：
| n |
| 2n+1 |
点评：本题主要考查 了数列的裂项求和方法的应用，解题的关键是对数列的通项公式的变形
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