試題分析:(1)由題設(shè)形式可以看出,題設(shè)中給出了關(guān)于數(shù)列a
n的面的一個方程,即一個遞推關(guān)系,所以應該對此遞推關(guān)系進行變形整理以發(fā)現(xiàn)其中所蘊含的規(guī)律,觀察發(fā)現(xiàn)若對方程兩邊取倒數(shù)則可以得到一個類似等差數(shù)列的形式,對其中參數(shù)進行討論,分類求其通項即可.
(2)由于本題中條件較少,解題思路不宜用綜合法直接分析出,故求解本題可以采取分析法的思路,由結(jié)論探究其成立的條件,再證明此條件成立,即可達到證明不等式的目的.
解:(1)∵
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(n≥2),
∴
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(n≥2),
當b=1時,
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(n≥2),
∴數(shù)列{
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}是以
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為首項,以1為公差的等差數(shù)列,
∴
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=1+(n﹣1)×1=n,即a
n=1,
當b>0,且b≠1時,
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(n≥2),
即數(shù)列{
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}是以
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=
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為首項,公比為
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的等比數(shù)列,
∴
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=
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×
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=
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,即a
n=
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,
∴數(shù)列{a
n}的通項公式是
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(2)證明:當b=1時,不等式顯然成立
當b>0,且b≠1時,a
n=
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,要證對于一切正整數(shù)n,2a
n≤b
n+1+1,只需證2×
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≤b
n+1+1,即證
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∵
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=
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=(b
n+1+1)×(b
n﹣1+b
n﹣2+…+b+1)
=(b
2n+b
2n﹣1+…+b
n+2+b
n+1)+(b
n﹣1+b
n﹣2+…+b+1)
=b
n[(b
n+b
n﹣1+…+b
2+b)+(
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+
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+…+
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)]
≥b
n(2+2+…+2)=2nb
n所以不等式成立,
綜上所述,對于一切正整數(shù)n,有2a
n≤b
n+1+1,
點評:本題考點是數(shù)列的遞推式,考查根據(jù)數(shù)列的遞推公式求數(shù)列的通項,研究數(shù)列的性質(zhì)的能力,本題中遞推關(guān)系的形式適合用取倒數(shù)法將所給的遞推關(guān)系轉(zhuǎn)化為有規(guī)律的形式,兩邊取倒數(shù),條件許可的情況下,使用此技巧可以使得解題思路呈現(xiàn)出來.數(shù)列中有請多成熟的規(guī)律,做題時要注意積累這些小技巧,在合適的情況下利用相關(guān)的技巧,可以簡化做題.在(2)的證明中,采取了分析法的來探究解題的思路,通過本題希望能進一步熟悉分析法證明問題的技巧.