题目内容
已知数列{an}中,a1=1,a2=3,其前n项和为Sn,且当n≥2时,an+1Sn-1-anSn=0.(Ⅰ)求证:数列{Sn}是等比数列;
(Ⅱ)求数列{an}的通项公式;
(Ⅲ)令

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【答案】分析:(1)由Sn与an的关系得an+1Sn-1-anSn=(Sn+1-Sn)Sn-1-(Sn-Sn-1)Sn=Sn+1Sn-1-Sn2整理得Sn2=Sn-1Sn+1s所以数列{Sn}是等比数列
(2)由(1)先求出Sn=4n-1接着当n≥2时,an=Sn-Sn-1=3×4n-2验证n=1也成立,可求出数列{an}的通项公式.
(3)把an的通项公式代入
得bn的通项公式求出Tn,利用其单调性与放缩法证明不等式
.
解答:(Ⅰ)证明:当n≥2时,
an+1Sn-1-anSn=(Sn+1-Sn)Sn-1-(Sn-Sn-1)Sn=Sn+1Sn-1-Sn2,
所以Sn2=Sn-1Sn+1(n≥2).
又由S1=1≠0,S2=4≠0,可推知对一切正整数n均有Sn≠0,
∴数列{Sn}是等比数列.
(Ⅱ)解:由(Ⅰ)知等比数列{Sn}的首项为1,公比为4,
∴Sn=4n-1.当n≥2时,an=Sn-Sn-1=3×4n-2,又a1=S1=1,
∴
(Ⅲ)证明:当n≥2时,an=3×4n-2,
此时
=
,
又
,
∴
.
当n≥2时,
=

=
.
又因为对任意的正整数n都有bn>0,所以Tn单调递增,即Tn≥T1,
∵
所以对于任意的正整数n,都有
成立.
点评:考查Sn与an的关系与分类讨论的思想,在这里求数列通项公式以及运用单调性与放缩法求和的对计算能力也有一定的要求.
(2)由(1)先求出Sn=4n-1接着当n≥2时,an=Sn-Sn-1=3×4n-2验证n=1也成立,可求出数列{an}的通项公式.
(3)把an的通项公式代入


解答:(Ⅰ)证明:当n≥2时,
an+1Sn-1-anSn=(Sn+1-Sn)Sn-1-(Sn-Sn-1)Sn=Sn+1Sn-1-Sn2,
所以Sn2=Sn-1Sn+1(n≥2).
又由S1=1≠0,S2=4≠0,可推知对一切正整数n均有Sn≠0,
∴数列{Sn}是等比数列.
(Ⅱ)解:由(Ⅰ)知等比数列{Sn}的首项为1,公比为4,
∴Sn=4n-1.当n≥2时,an=Sn-Sn-1=3×4n-2,又a1=S1=1,
∴

(Ⅲ)证明:当n≥2时,an=3×4n-2,
此时


又

∴

当n≥2时,
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=

又因为对任意的正整数n都有bn>0,所以Tn单调递增,即Tn≥T1,
∵

所以对于任意的正整数n,都有

点评:考查Sn与an的关系与分类讨论的思想,在这里求数列通项公式以及运用单调性与放缩法求和的对计算能力也有一定的要求.

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