题目内容
已知数列
,首项a 1 =3且2a n+1="S" n?S n-1 (n≥2).
(1)求证:{
}是等差数列,并求公差;
(2)求{a n }的通项公式;
(3)数列{an }中是否存在自然数k0,使得当自然数k≥k 0时使不等式a k>a k+1对任意大于等于k的自然数都成立,若存在求出最小的k值,否则请说明理由.
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(1)求证:{
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(2)求{a n }的通项公式;
(3)数列{an }中是否存在自然数k0,使得当自然数k≥k 0时使不等式a k>a k+1对任意大于等于k的自然数都成立,若存在求出最小的k值,否则请说明理由.
(1)
(2)
(3)3
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(2)
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(3)3
试题分析:解:⑴由已知当
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⑶
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点评:解决的关键是通过数列的递推关系来分析得到证明等差数列,同事借助于关系式得到{a n },然后借助于不等式来得到参数的范围,属于基础题。
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