题目内容
已知数列| 1 |
| 1•2 |
| 1 |
| 2•3 |
| 1 |
| 3•4 |
| 1 |
| n(n+1) |
分析：根据题意，采用裂项求和法可以直接求出Sn的值．
解答：解：Sn=
+
+
+…+
=(1-
)+(
-
)+(
-
)+…+(
-
)
=1-
=
．
故答案：
．
| 1 |
| 1•2 |
| 1 |
| 2•3 |
| 1 |
| 3×4 |
| 1 |
| n(n+1) |
=(1-
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| n |
| 1 |
| n+1 |
=1-
| 1 |
| n+1 |
| n |
| n+1 |
故答案：
| n |
| n+1 |
点评：本题考查裂项求和法，解题时要认真审题，仔细求解．
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