题目内容
数列
，
，
，…，
，…的前n项和为（ ）
| 1 |
| 1×3 |
| 1 |
| 3×5 |
| 1 |
| 5×7 |
| 1 |
| (2n-1)(2n+1) |
分析：由题意可得，an=
=
(
-
)，利用裂项即可求解
| 1 |
| (2n-1)(2n+1) |
| 1 |
| 2 |
| 1 |
| 2n-1 |
| 1 |
| 2n+1 |
解答：解：由题意可得，an=
=
(
-
)
∴Sn=
(1-
+
-
+…+
-
)
=
(1-
)=
故选B
| 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 |
故选B
点评：本题主要考查了裂项求解数列的和，注意本题中裂项的规律
=
(
-
)中的
容易漏掉
| 1 |
| k(k+m) |
| 1 |
| m |
| 1 |
| k |
| 1 |
| k+m |
| 1 |
| m |
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