【答案】
分析:(Ⅰ)由S
n=2a
n+
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×(-1)
n-
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,n=1,2,3,…,再寫一式,兩式相減整理可得a
n=2a
n-1+3×(-1)
n-1
(Ⅱ)由(Ⅰ)令b
n=(-1)
na
n得b
n=-2b
n-1-3,構(gòu)造新數(shù)列b
n+1是等比數(shù)列,從而可求數(shù)列{a
n}的通項公式;
(Ⅲ)由
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,∴S
2k-1=2
2k-1,S
2k=2
2k-1,再進(jìn)行分組求和,利用等比數(shù)列的求和公式可證.
解答:解:(Ⅰ)由S
n=2a
n+
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×(-1)
n-
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,n=1,2,3,…,①
得S
n-1=2a
n-1+
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×(-1)
n-1-
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,n=2,3,…,②…(1分)
將①和②相減得:
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,n=2,3,…,…(2分)
整理得:a
n=2a
n-1+3×(-1)
n-1,n=2,3,…. …(3分)
(Ⅱ)在已知條件中取n=1得,a
1=2a
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-
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,∴a
1═2.…(4分)
∵a
n=2a
n-1+3×(-1)
n-1,∴(-1)
na
n=-2(-1)
n-1a
n-1-3,
∴令b
n=(-1)
na
n得b
n=-2b
n-1-3,n=2,3,….…(5分)
∴b
n+1+1=-2(b
n+1),n=1,2,3,…,
∵b
1+1=-1≠0,∴b
n+1=(-1)×(-2)
n-1,n=1,2,3,…,
∴a
n=2
n-1+(-1)
n-1. …(7分)
(Ⅲ)∵
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,∴S
2k-1=2
2k-1,S
2k=2
2k-1. …(8分)
∴
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+
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+…+
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=
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<
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. …(10分)
同理
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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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,n∈N
*. …(12分)
點評:本題主要考查數(shù)列通項公式的求解,考查數(shù)列與不等式的綜合,有一定的難度.