【答案】
分析:(1)將b
n=a
n-1代入2a
n=1+a
na
n+1,可得b
n的遞推關(guān)系式,整理變形可得
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,由等差數(shù)列的定義可得
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為等差數(shù)列,故可求其通項公式,進而求出b
n.
(2)結(jié)合(1)中的結(jié)論,寫出T
n+1-T
n的表達式,利用放縮法證明該差大于0即可.
(3)利用疊加法把
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轉(zhuǎn)化為
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+
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+…+T
2+T
1+S
1的形式,再結(jié)合(2)中的結(jié)論,利用T
n的單調(diào)性證明不等式.
解答:解:(1)由b
n=a
n-1,得a
n=b
n+1,代入2a
n=1+a
na
n+1,
得2(b
n+1)=1+(b
n+1)(b
n+1+1),
∴b
nb
n+1+b
n+1-b
n=0,從而有
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,
∵b
1=a
1-1=2-1=1,
∴
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是首項為1,公差為1的等差數(shù)列,
∴
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,即
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;(5分)
(2)∵
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,
∴
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,
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,
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,
∴T
n+1>T
n;(10分)
(3)∵n≥2,
∴
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=
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+
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+…+T
2+T
1+S
1.
由(2)知
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≥
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≥…≥T
2≥T
1≥S
1,
∵
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,
∴
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=
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+
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+…+T
2+T
1+S
1≥(n-1)T
2+T
1+S
1=
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=
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.(16分)
點評:本題考查了數(shù)列和不等式的綜合應用,應用了構(gòu)造法、放縮法、疊加法等數(shù)學思想方法,難度較大.