题目内容
对于正整数n,求证:1+
+
+…+
>2(
-1)
| 1 | ||
|
| 1 | ||
|
| 1 | ||
|
| n |
考点:不等式的证明
专题:不等式
分析:不等式右边有个2,所以将不等式左边提取2变成:2(
+
+
+…+
)>2(
+
+
+…+
+
)=2(
-1+
-
+
-
+…
-
+
)=2(
-1)+
>2(
-1).
| 1 |
| 2 |
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2
|
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2
|
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2
|
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|
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2
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| 3 |
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| 4 |
| 3 |
| n |
| n-1 |
| 1 | ||
2
|
| n |
| 1 | ||
|
| n |
解答:
证明:1+
+
+…+
=2(
+
+
+…+
)>2(
+
+
+…+
+
)=2(
-1+
-
+
-
+…+
-
+
)=2(
-1)+
>2(
-1).
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2
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2
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2
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2
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| n |
| n-1 |
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2
|
| n |
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|
| n |
点评:考查放缩法证明不等式的方法,想着用放缩法的原因是:不等式右边是2(
-1),所以不等式左边提取2,并想法消去中间项,这样便可以想着用放缩法.
| n |
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