题目内容
（2011•蓝山县模拟）S=
+
+…+
=
．
| 1 | ||
1+
|
| 1 | ||||
|
| 1 | ||||
|
| ||
| 2 |
| ||
| 2 |
分析：由题意可得，S=
+
+…+
=
+
+…+
，利用裂项相消可求和
| 1 | ||
1+
|
| 1 | ||||
|
| 1 | ||||
|
1-
| ||
| -2 |
| ||||
| -2 |
| ||||
| -2 |
解答：解：由题意可得，S=
+
+…+
=
+
+…+
=-
（1-
+
-
+…+
-
）
=-
（1-
）=
故答案为
| 1 | ||
1+
|
| 1 | ||||
|
| 1 | ||||
|
=
1-
| ||
| -2 |
| ||||
| -2 |
| ||||
| -2 |
=-
| 1 |
| 2 |
| 3 |
| 3 |
| 5 |
| 2009 |
| 2011 |
=-
| 1 |
| 2 |
| 2011 |
| ||
| 2 |
故答案为
| ||
| 2 |
点评：本题主要考查了裂项求解数列的和，解题的关键是在所求的每项中乘以分母的有理化因式．
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