题目内容
已知各项均为正数的数列{a
}满足a
=2a
+a
a
,且a
+a
=2a
+4,其中n∈N
.
(Ⅰ)若b
=
,求数列{b
}的通项公式;
(Ⅱ)证明:
+
+…+
>
(n≥2).
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(Ⅰ)若b
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(Ⅱ)证明:
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(1)b
=
(n∈N
)
(2)构造函数借助于函数的最值来证明不等式。
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(2)构造函数借助于函数的最值来证明不等式。
试题分析:解:(Ⅰ)因为a
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又a
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所以数列
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由
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从而,数列{a
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(Ⅱ)构造函数f(x)=
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则f′(x)=
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当0<x<b
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所以f(x)的最大值是f(b
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即
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所以
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令x=
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所以
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即
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点评:解决的关键是能利用等比数列来求解通项公式,同时能结合导数来拍脑袋函数单调性,以及求解函数的最值,同时证明不等式,属于中档题。
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