【答案】
分析:(1)把n•a
n+1=(n+2)S
n代入a
n+1=S
n+1-S
n中化簡整理得
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=2
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.進而可推斷數(shù)列
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是公比為2的等比數(shù)列.
(2)根據(jù)又
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求的首項,進而根據(jù)等比數(shù)列的通項公式求的數(shù)列
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的通項公式,進而求的S
n關(guān)于n的表達式.
(3)把(2)中求的S
n關(guān)于n的表達式代入
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中,結(jié)果大于1,進而可判斷{S
n}為遞增數(shù)列,進而可知存在N
=8,對所有n>N
有S
n>2007恒成立.
解答:解:(1)證明:∵a
n+1=S
n+1-S
n,
由已知a
n+1=
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S
n,∴
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S
n=S
n+1-S
n,
(n+2)S
n=nS
n+1-nS
n,2(n+1)S
n=nS
n+1,
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=2
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.又
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=
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=1,
∴
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是以1為首項,2為公比的等比數(shù)列
(2)∵
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=1•2
n-1=2
n-1,∴S
n=n•2
n-1.
(3)猜測:存在N
=8,當n>8時有S
n>2007恒成立
∵
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=
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=

>1,
∴{S
n}為遞增數(shù)列,
∴存在N
=8,對所有n>N
有S
n>2007恒成立
點評:本題主要考查了等比關(guān)系的確定和數(shù)列與不等式問題的綜合考查.數(shù)列與函數(shù)、不等式、對數(shù)等問題的綜合考查是近幾年高考的熱點問題.