题目内容
求证:
+
+…+
<
.
| 1 |
| 2×3 |
| 1 |
| 4×5 |
| 1 |
| (2n)(2n+1) |
| 1 |
| 3 |
考点:反证法与放缩法
专题:选作题,反证法
分析:利用(2n)(2n+1)>(2n-1)(2n+1),可得
<
=
(
-
),即可证明结论.
| 1 |
| (2n)(2n+1) |
| 1 |
| (2n-1)(2n+1) |
| 1 |
| 2 |
| 1 |
| 2n-1 |
| 1 |
| 2n+1 |
解答:
证明:∵(2n)(2n+1)>(2n-1)(2n+1),
∴
<
=
(
-
),
∴
+
+…+
<
+
(+
-
+…+
-
)=
+
(
-
)<
.
即
+
+…+
<
.
∴
| 1 |
| (2n)(2n+1) |
| 1 |
| (2n-1)(2n+1) |
| 1 |
| 2 |
| 1 |
| 2n-1 |
| 1 |
| 2n+1 |
∴
| 1 |
| 2×3 |
| 1 |
| 4×5 |
| 1 |
| (2n)(2n+1) |
| 1 |
| 6 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 5 |
| 1 |
| 2n-1 |
| 1 |
| 2n+1 |
| 1 |
| 6 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 2n+1 |
| 1 |
| 3 |
即
| 1 |
| 2×3 |
| 1 |
| 4×5 |
| 1 |
| (2n)(2n+1) |
| 1 |
| 3 |
点评:本题考查放缩法,考查学生分析解决问题的能力,利用
<
=
(
-
)是关键.
| 1 |
| (2n)(2n+1) |
| 1 |
| (2n-1)(2n+1) |
| 1 |
| 2 |
| 1 |
| 2n-1 |
| 1 |
| 2n+1 |
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