解:(1)不等式x
2-x<(2n-1)x 即x(x-2n)<0,解得:0<x<2n,其中整數(shù)有2n-1個,
故 a
n=2n-1.
(2)由(1)知
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,∴S
m=m
2,S
p=p
2,S
k=k
2.
由
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=
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=
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≥
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=0,
即
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≥
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.
(3)結(jié)論成立,證明如下:
設(shè)等差數(shù)列{a
n}的首項為a
1,公差為d,則
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,
∵
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=
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,
把m+p=2k代入上式化簡得S
m+S
p-2S
k=
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≥0,…16分.
∴S
m+S
p≥2S
k.
又 S
m•S
p =
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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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=
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,故
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+
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≥
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成立.
分析:(1)由題意知數(shù)列{a
n}的通項是關(guān)于x的不等式的解集中整數(shù)的個數(shù),題目首先應(yīng)該解不等式,從不等式的解集中得到整數(shù)的個數(shù),得到數(shù)列的通項,用等差數(shù)列的定義來驗證.
(2)根據(jù)前面結(jié)果寫出要用的前幾項的和,從不等式的一側(cè)入手,利用均值不等式得到要求的結(jié)論.
(3)本題是對上一問的延伸,方法和前面的類似,但題目所給的一般的各項均為正數(shù)的等差數(shù)列在整理時增加了難度,題目絕大部分工作是算式的整理,注意不能出錯.
點評:本題沒有具體的數(shù)字運算但運算量非常大,它考查的是等差數(shù)列和等比數(shù)列的性質(zhì),基本不等式,實際上這類問題比具體的數(shù)字運算要困難,是幾個知識點結(jié)合起來的綜合問題,屬于中檔題.