已知Sn是正项数列{an}的前n项和,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,求数列{bn}的前n项和Tn;(3)若bn≤14m2-m-12

题目简介

已知Sn是正项数列{an}的前n项和,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,求数列{bn}的前n项和Tn;(3)若bn≤14m2-m-12

题目详情

已知Sn是正项数列{an}的前n项和,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
,求数列{bn}的前n项和Tn
(3)若bn
1
4
m2-m-
1
2
对一切正整数n恒成立,求实数m的取值范围.
题型:解答题难度:中档来源:不详

答案

(1)当 n≥2时,Sn=class="stub"1
4
(an+1)2
,Sn-1=class="stub"1
4
(an-1+1)2

两式相减,整理得(an+an-1)(an-an-1-2)=0,由于数列{an}是正项数列,∴an-an-1=2,又a1=1,所以数列{an}是首项a1=1,d=2的等差数列,an=2n-1;
(2)bn=
an
2n
=class="stub"2n-1
2n
Tn=class="stub"1
2
+class="stub"3
22
++class="stub"2n-1
2n
class="stub"1
2
Tn=class="stub"1
22
+class="stub"3
23
++class="stub"2n-1
2n+1

相减化简得Tn=3-class="stub"2n+3
2n

(3)∵bn+1-bn=class="stub"3-2n
2n+1

当n=1,b2>b1,当n≥2,bn+1<bn,故当n=2时,b2取到最大值class="stub"3
4

bn≤class="stub"1
4
m2-m-class="stub"1
2
对一切正整数n恒成立,即class="stub"3
4
≤class="stub"1
4
m2-m-class="stub"1
2

解得m≤-1或m≥5

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