题目内容
证明:
+
+…+
<
,n∈Z*.
| 1 |
| 12 |
| 1 |
| 22 |
| 1 |
| n2 |
| 7 |
| 4 |
考点:数学归纳法
专题:证明题,点列、递归数列与数学归纳法
分析:利用
<
=
(
-
),即可证明结论.
| 1 |
| n2 |
| 1 |
| n2-1 |
| 1 |
| 2 |
| 1 |
| n-1 |
| 1 |
| n+1 |
解答:
证明:∵
<
=
(
-
)
∴
+
+…+
<1+
(1-
+
-
+
-
+…+
-
)=1+
(1+
-
-
)<
.
| 1 |
| n2 |
| 1 |
| n2-1 |
| 1 |
| 2 |
| 1 |
| n-1 |
| 1 |
| n+1 |
∴
| 1 |
| 12 |
| 1 |
| 22 |
| 1 |
| n2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 3 |
| 1 |
| 5 |
| 1 |
| n-1 |
| 1 |
| n+1 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| n |
| 1 |
| n+1 |
| 7 |
| 4 |
点评:本题考查不等式的证明,考查放缩法的运用,利用
<
=
(
-
)是关键.
| 1 |
| n2 |
| 1 |
| n2-1 |
| 1 |
| 2 |
| 1 |
| n-1 |
| 1 |
| n+1 |
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