题目内容
若n是大于1的自然数,求证
+
+…+
>
-
.
| 1 |
| 22 |
| 1 |
| 32 |
| 1 |
| n2 |
| 1 |
| 2 |
| 1 |
| n+1 |
分析:利用
>
,可得
>
-
,再叠加,即可证得结论
| 1 |
| n2 |
| 1 |
| n(n+1) |
| 1 |
| n2 |
| 1 |
| n |
| 1 |
| n+1 |
解答:证明:∵
>
∴
>
-
∴
+
+…+
>
-
+
-
+…+
-
∴
+
+…+
>
-
| 1 |
| n2 |
| 1 |
| n(n+1) |
∴
| 1 |
| n2 |
| 1 |
| n |
| 1 |
| n+1 |
∴
| 1 |
| 22 |
| 1 |
| 32 |
| 1 |
| n2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| n |
| 1 |
| n+1 |
∴
| 1 |
| 22 |
| 1 |
| 32 |
| 1 |
| n2 |
| 1 |
| 2 |
| 1 |
| n+1 |
点评:本题重点考查不等式的证明,考查放缩法的运用,解题的关键是利用
>
,可得
>
-
,
| 1 |
| n2 |
| 1 |
| n(n+1) |
| 1 |
| n2 |
| 1 |
| n |
| 1 |
| n+1 |
练习册系列答案
相关题目