题目内容
证明：
≤
+
+…+
＜
（n为正整数）．
| 1 |
| 3 |
| 1 |
| 1×3 |
| 1 |
| 3×5 |
| 1 |
| (2n-1)(2n+1) |
| 1 |
| 2 |
证明：
+
+…+
=
（1-
）+
（
-
）+…+
（
-
）
=
（1-
+
-
+…+
-
）
=
（1-
）
=
•
=
，
∵
≤
＜
，（n为正整数，n=1时
最小），
∴
≤
＜
，
∴
≤
+
+…+
＜
（n为正整数）．
| 1 |
| 1×3 |
| 1 |
| 3×5 |
| 1 |
| (2n-1)(2n+1) |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 5 |
| 1 |
| 2 |
| 1 |
| 2n-1 |
| 1 |
| 2n+1 |
=
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 5 |
| 1 |
| 2n-1 |
| 1 |
| 2n+1 |
=
| 1 |
| 2 |
| 1 |
| 2n+1 |
=
| 1 |
| 2 |
| 2n |
| 2n+1 |
=
| n |
| 2n+1 |
∵
| 1 |
| 2×1+1 |
| n |
| 2n+1 |
| n |
| 2n |
| n |
| 2n+1 |
∴
| 1 |
| 3 |
| n |
| 2n+1 |
| 1 |
| 2 |
∴
| 1 |
| 3 |
| 1 |
| 1×3 |
| 1 |
| 3×5 |
| 1 |
| (2n-1)(2n+1) |
| 1 |
| 2 |
练习册系列答案
相关题目
