【答案】
分析:(Ⅰ)由
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,得當(dāng)n≥2時(shí),S
n=2S
n-1+n,兩式相減得,a
n+1=2a
n+1,構(gòu)造等比數(shù)列{a
n+1}并求其通項(xiàng)公式,再求出數(shù)列{a
n}的通項(xiàng)公式.
(Ⅱ)b
n=
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=
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=
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,利用錯(cuò)位相消法求和.
解答:解:(Ⅰ)∵
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當(dāng)n≥2時(shí),S
n=2S
n-1+n,兩式相減得,
a
n+1=2a
n+1,兩邊加上1得出a
n+1+1=2(a
n+1),
又S
2=2S
1+1,a
1=S
1=1,∴a
2=3,a
2+1=2(a
1+1)
所以數(shù)列{a
n+1}是公比為2的等比數(shù)列,首項(xiàng)a
1+1=2,
數(shù)列{a
n+1}的通項(xiàng)公式為a
n+1=2•2
n-1=2
n,
∴a
n=2
n-1
(Ⅱ)∵a
n=2
n-1,
∴b
n=
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=
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=
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T
n=
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T
n=
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兩式相減得
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T
n=
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T
n=2(
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)=2
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<2.
點(diǎn)評(píng):本題主要考查數(shù)列通項(xiàng)公式求解:利用了a
n與Sn關(guān)系以及構(gòu)造法.形如a
n+1=pa
n+q遞推數(shù)列,這種類(lèi)型可轉(zhuǎn)化為a
n+1+m=4(a
n+m)構(gòu)造等比數(shù)列求解.還考查錯(cuò)位相消法求和.