【答案】
分析:(Ⅰ)先求等比數(shù)列{a
n}的前n項(xiàng)和S
n,再表達(dá)出
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,故可證;
(II)先求出b
n,再進(jìn)一步變形,判斷出
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是等差數(shù)列,根據(jù)等差數(shù)列的通項(xiàng)公式求出{b
n}的通項(xiàng)公式;
(III)先求出C
n,再由錯(cuò)位相減法求出該數(shù)列的前n項(xiàng)和為T(mén)
n.
解答:解:(Ⅰ)證明:
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而
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所以S
n=(1+λ)-λa
n(4分)
(Ⅱ)
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,∴
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,∴
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,(6分)
∴
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是首項(xiàng)為
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,公差為1的等差數(shù)列,
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,即
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.(8分)
(Ⅲ)λ=1時(shí),
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,∴
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(9分)
∴
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∴
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相減得∴
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∴
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,(12分)
又因?yàn)?img src="http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182335200921008/SYS201310241823352009210021_DA/17.png">,∴T
n單調(diào)遞增,
∴T
n≥T
2=2,故當(dāng)n≥2時(shí),2≤T
n<4.(13分)
點(diǎn)評(píng):本題是數(shù)列的綜合題,考查等差數(shù)列、等比數(shù)列的通項(xiàng)公式,涉及了錯(cuò)位相減法求數(shù)列的前n項(xiàng)和,考查了分析問(wèn)題和解決問(wèn)題的能力.