(1)解:∵
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,∴
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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)證明:由
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,
∴
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=
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,
∴
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,即a
n-a
n+1=a
na
n+1,
∴
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∴數列{
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}是以4為首項,1為公差的等差數列.
∴
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,則
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,
∴
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;
(3)解:由
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,
∴S
n=a
1a
2+a
2a
3+…+a
na
n+1=
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=
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=
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.
∴
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,
要使4λS
n<b
n恒成立,只需(λ-1)n
2+(3λ-6)n-8<0恒成立,
設f(n)=(λ-1)n
2+3(λ-2)n-8
當λ=1時,f(n)=-3n-8<0恒成立,
當λ>1時,由二次函數的性質知f(n)不滿足對于任意n∈N
*恒成立,
當λ<l時,對稱軸n=
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f(n)在[1,+∞)為單調遞減函數.
只需f(1)=(λ-1)n
2+(3λ-6)n-8=(λ-1)+(3λ-6)-8=4λ-15<0
∴
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,∴λ≤1時4λS
n<b
n恒成立.
綜上知:λ≤1時,4λS
n<b
n恒成立.
分析:(1)由給出的
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,循環(huán)代入a
n+b
n=1和
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可求解a
2,a
3;
(2)由a
n+b
n=1得a
n+1+b
n+1=1,結合
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,去掉b
n與b
n+1得到a
n+1與a
n的關系式,整理變形后可證得數列{
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}是以4為首項,1為公差的等差數列,求出其通項公式后即可求得數列{a
n}和{ b
n}的通項公式;
(3)首先利用裂項求和求出S
n,代入4λS
n<b
n,通過對λ分類討論,結合二次函數的最值求使4λS
n<b
n恒成立的實數λ的值.
點評:本題考查了等差、等比數列的通項公式,考查了數列的裂項求和,考查了數列的函數特性,訓練了恒成立問題的求解方法,解答過程中注意分類討論的數學思想,屬中檔題.