已知正项数列{an}的前n和为Sn,且Sn是14与(an+1)2的等比中项.(1)求证:数列{an}是等差数列;(2)若bn=an2n,数列{bn}的前n项和为Tn,求Tn;(3)在(2)的条件下,是

题目简介

已知正项数列{an}的前n和为Sn,且Sn是14与(an+1)2的等比中项.(1)求证:数列{an}是等差数列;(2)若bn=an2n,数列{bn}的前n项和为Tn,求Tn;(3)在(2)的条件下,是

题目详情

已知正项数列{an}的前n和为Sn,且
Sn
1
4
与(an+1)2的等比中项.
(1)求证:数列{an}是等差数列;
(2)若bn=
an
2n
,数列{bn}的前n项和为Tn,求Tn
(3)在(2)的条件下,是否存在常数λ,使得数列{
Tn
an+2
}
为等比数列?若存在,试求出λ;若不存在,说明理由.
题型:解答题难度:中档来源:不详

答案

(1)∵Sn=class="stub"1
4
(an+1)2
,∴a1=S1=class="stub"1
4
(a1+1)2
,∴a1=1(an>0)
当n≥2时,an=Sn-Sn-1=class="stub"1
4
(an+1)2-class="stub"1
4
(an-1+1)2
,∴(an+an-1)(an-an-1-2)=0
∵an>0,
∴an-an-1=2,
∴{an}为等差数列.(4')
(2)由(1)知,{an}是以1为首项,2为公差的等差数列,
∴an=2n-1
bn=class="stub"2n-1
2n
,①
Tn=class="stub"1
2
+class="stub"3
22
+…+class="stub"2n-1
2n
,①
class="stub"1
2
Tn=    class="stub"1
22
+class="stub"3
23
+class="stub"5
24
+…+class="stub"2n-3
2n
+class="stub"2n-1
2n

①-②得:class="stub"1
2
Tn=class="stub"1
2
+2(class="stub"1
22
+class="stub"1
23
+class="stub"1
24
+class="stub"1
2n
)-class="stub"2n-1
2n+1

Tn=3-class="stub"2n-3
2n
(9')
(3)∵
Tn
an+2
=(3-class="stub"2n+3
2n
+λ)class="stub"1
2n+3
=class="stub"3+λ
2n+3
-class="stub"1
2n

易知,当λ=-3时,数列{
Tn
an+2
}
为等比数列.(13')

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