题目内容
| 1 |
| 1•2 |
| 1 |
| 2•3 |
| 1 |
| n(n+1) |
| n |
| n+1 |
| n |
| n+1 |
分析:结合数列的特点,
=
-
,可考虑利用裂项求和进行求解即可
| 1 |
| n(n+1) |
| 1 |
| n |
| 1 |
| n+1 |
解答:解:∵
=
-
∴
+
+…
=1-
+
-
+…+
-
=1-
=
故答案为:
| 1 |
| n(n+1) |
| 1 |
| n |
| 1 |
| n+1 |
∴
| 1 |
| 1•2 |
| 1 |
| 2•3 |
| 1 |
| n(n+1) |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| n |
| 1 |
| n+1 |
=1-
| 1 |
| n+1 |
| n |
| n+1 |
故答案为:
| n |
| n+1 |
点评:本题主要考查了数列求和的裂项求和,属于基本方法的应用,属于基础试题
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