【答案】
分析:(1)根據(jù)
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,求出
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,和
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,令n=1,2,3即可求得b
1,b
2,b
3,b
4;
(2)根據(jù)
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,進(jìn)行變形得到
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,構(gòu)造等差數(shù)列{
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},并求出其通項(xiàng),進(jìn)而可求出數(shù)列{b
n}的通項(xiàng)公式;
(3)根據(jù)(2)結(jié)果,可以求出數(shù)列{a
n}的通項(xiàng)公式,然后利用裂項(xiàng)相消法求S
n,構(gòu)造函數(shù)f(n)=(a-1)n
2+(3a-6)n-8,轉(zhuǎn)化為求函數(shù)f(n)的最值問題,可求實(shí)數(shù)a的取值范圍.
解答:解:(1)∵
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∴
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,
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,
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,
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,
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(2)∵
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∴
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∴數(shù)列{
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}是以-4為首項(xiàng),-1為公差的等差數(shù)列
∴
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∴
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;
(3)
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,
∴
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∴
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由條件可知(a-1)n
2+(3a-6)n-8<0恒成立即可滿足條件,
設(shè)f(n)=(a-1)n
2+(3a-6)n-8
當(dāng)a=1時(shí),f(n)=-3n-8<0恒成立
當(dāng)a>1時(shí),由二次函數(shù)的性質(zhì)知不可能成立
當(dāng)a<1時(shí),對(duì)稱軸
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f(n)在(1,+∞)為單調(diào)遞減函數(shù).
f(1)=(a-1)n
2+(3a-6)n-8=(a-1)+(3a-6)-8=4a-15<0
∴
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∴a<1時(shí)4aS
n<b恒成立
綜上知:a≤1時(shí),4aS
n<b恒成立.
點(diǎn)評(píng):此題是個(gè)難題.考查根據(jù)數(shù)列的遞推公式利用構(gòu)造法求數(shù)列的通項(xiàng)公式,及數(shù)列的求和問題,題目綜合性強(qiáng),特別是問題(3)的設(shè)置,數(shù)列與不等式恒成立問題結(jié)合起來(lái),能有效考查學(xué)生的邏輯思維能力,體現(xiàn)了轉(zhuǎn)化的思想和分類討論的思想.