【答案】
分析:(I)根據(jù)S
n,是na
n,a
n的等差中項,得出na
n+a
n=2S
n,(n+1)a
n=2S
n,又2S
n-2S
n-1=2a
n∴a
n=
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a
n-1,求得a
n=n.得出
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=
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=
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;
(II)由于數(shù)列{
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}的前n項和為T
n∴
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,(n+1)T
n+1-nT
n=(n+1)(T
n+
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)-nT
n=1+T
n,從而得出(n+1)T
n+1-nT
n=1+T
n(III)對于存在性問題,可先假設(shè)存在,即假設(shè)存在數(shù)列{b
n},再利用條件,求出b
n,若出現(xiàn)矛盾,則說明假設(shè)不成立,即不存在;否則存在.
解答:解:(I)∵S
n,是na
n,a
n的等差中項
∴na
n+a
n=2S
n,
∴(n+1)a
n=2S
n,
∵2S
n-2S
n-1=2a
n∴(n+1)a
n-na
n-1=2a
n∴a
n=
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a
n-1∴a
n=n.
∴
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=
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=
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;
(II)∵數(shù)列{
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}的前n項和為T
n∴
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,
∴(n+1)T
n+1-nT
n=(n+1)(T
n+
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)-nT
n=1+T
n∴(n+1)T
n+1-nT
n=1+T
n(III)假設(shè)存在數(shù)列{b
n},使Pn=(b
n+1)T
n-b
n,
當(dāng)n=2時,有:P
2=(b
2+1)T
2-b
2,
即:1+1+
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═(b
2+1)(1+
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)-b
2,
∴b
2=4,
當(dāng)n=3時,有:P
3=(b
3+1)T
3-b
3,
即:1+1+

+1+
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+

=(b
3+1)(1+
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+
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)-b
3,
∴b
3=3,
…
依此類推,存在數(shù)列{b
n},b
n=5-n.
使得Pn=(b
n+1)T
n-b
n.
點評:本題考查數(shù)列的應(yīng)用,數(shù)列的極限.注意(Ⅲ)的處理存在性問題的一般方法,首先假設(shè)存在,進(jìn)而根據(jù)題意、結(jié)合有關(guān)性質(zhì),化簡、轉(zhuǎn)化、計算,最后得到結(jié)論.