已知数列{an}是等差数列,a3=10,a6=22,数列{bn}的前n项和是Tn,且Tn+13bn=1.(Ⅰ)求数列{an}的通项公式;(Ⅱ)求证:数列{bn}是等比数列;(Ⅲ)记cn=an•bn,求

题目简介

已知数列{an}是等差数列,a3=10,a6=22,数列{bn}的前n项和是Tn,且Tn+13bn=1.(Ⅰ)求数列{an}的通项公式;(Ⅱ)求证:数列{bn}是等比数列;(Ⅲ)记cn=an•bn,求

题目详情

已知数列{an}是等差数列,a3=10,a6=22,数列{bn}的前n项和是Tn,且Tn+
1
3
bn=1

(Ⅰ)求数列{an}的通项公式;
(Ⅱ)求证:数列{bn}是等比数列;
(Ⅲ)记cn=an•bn,求证:cn+1<cn
题型:解答题难度:中档来源:不详

答案

(I)∵数列{an}是等差数列,a3=10,a6=22,
a1+2d=10
a1+5d=22.
解得 a1=2,d=4.
∴an=2+(n-1)×4=4n-2.…(4分)
(II)证明:由于Tn=1-class="stub"1
3
bn
,①
令n=1,得b1=1-class="stub"1
3
b1
,解得b1=class="stub"3
4

当n≥2时,Tn-1=1-class="stub"1
3
bn-1

①-②得bn=class="stub"1
3
bn-1-class="stub"1
3
bn

bn=class="stub"1
4
bn-1

b1=class="stub"3
4
≠0
,∴
bn
bn-1
=class="stub"1
4

∴数列{bn}是以class="stub"3
4
为首项,class="stub"1
4
为公比的等比数列.…(9分)
(III)证明:由(II)可得bn=class="stub"3
4n
.…(9分)   
cn=anbn=
3(4n-2)
4n
…(10分)
cn+1-cn=
3[4(n+1)-2]
4n+1
-
3(4n-2)
4n
=class="stub"30-36n
4n+1

∵n≥1,故cn+1-cn<0,
∴cn+1<cn.…(13分)

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