已知數(shù)列
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滿足:
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(1)求
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的值;
(2)求證:數(shù)列

是等比數(shù)列;
(3)令
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(
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),如果對任意
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,都有
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,求實(shí)數(shù)

的取值范圍.
(1)
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;(2)
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是以
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為首相
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為公比的等比數(shù)列;
(3)
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試題分析:(1)利用賦值法,令
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可求
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;
(2)將等式寫到
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,再將得到的式子與已知等式聯(lián)立,兩式再相減,根據(jù)等比數(shù)列的定
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,可證明
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是以
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為首相

為公比的等比數(shù)列;
(3)由(2)可寫出
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,利用數(shù)列的單調(diào)性當(dāng)

時,
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,當(dāng)
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時,
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,因此,數(shù)列
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的最大值為
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,則
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可解的
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的范圍.
試題解析:(1)
(2)由題可知:
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①
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②
②-①可得
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即:
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,又
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∴數(shù)列
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是以
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為首項,以

為公比的等比數(shù)列
(3)由(2)可得
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,
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由
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可得
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由

可得
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,所以
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故
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有最大值
所以,對任意
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,有
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如果對任意

,都有
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,即
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成立,
則
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,故有:
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,解得
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或
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∴實(shí)數(shù)

的取值范圍是
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
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