等差数列{an}中a3=7,a1+a2+a3=12,记Sn为{an}的前n项和,令bn=anan+1,数列{1bn}的前n项和为Tn.(1)求an和Sn;(2)求证:Tn<13;(3)是否存在正整数m

题目简介

等差数列{an}中a3=7,a1+a2+a3=12,记Sn为{an}的前n项和,令bn=anan+1,数列{1bn}的前n项和为Tn.(1)求an和Sn;(2)求证:Tn<13;(3)是否存在正整数m

题目详情

等差数列{ an}中a3=7,a1+a2+a3=12,记Sn为{an}的前n项和,令bn=anan+1,数列{
1
bn
}的前n项和为Tn
(1)求an和Sn
(2)求证:Tn
1
3

(3)是否存在正整数m,n,且1<m<n,使得T1,Tm,Tn成等比数列?若存在,求出m,n的值,若不存在,说明理由.
题型:解答题难度:中档来源:不详

答案

(1)设数列{an}的公差为d,
由a3=a1+2d=7,a1+a2+a3=3a1+3d=12,
解得a1=1,d=3,
∴an=3n-2,
Sn=n+
n(n-1)
2
×3
=
3n2-n
2

(2)∵bn=anan+1=(3n-2)(3n+1),
class="stub"1
bn
=class="stub"1
(3n-2)(3n+1)
=class="stub"1
3
(class="stub"1
3n-2
-class="stub"1
3n+1
)

Tn=class="stub"1
3
(1-class="stub"1
4
+class="stub"1
4
-class="stub"1
7
+class="stub"1
7
-class="stub"1
11
+…+
class="stub"1
3n-5
-class="stub"1
3n-2
+class="stub"1
3n-2
-class="stub"1
3n+1
)

=class="stub"1
3
(1-class="stub"1
3n+1
)<class="stub"1
3

(3)由(2)知,Tn=class="stub"n
3n+1
,∴T1=class="stub"1
4
Tm=class="stub"m
3m+1
Tn=class="stub"n
3n+1

∵T1,Tm,Tn成等比数列,
(class="stub"m
3m+1
)
2
=class="stub"1
4
×class="stub"n
3n+1

class="stub"6m+1
m2
=class="stub"3n+4
n

当m=1时,7=class="stub"3n+4
n
,n=1,不合题意;
当m=2时,class="stub"13
4
=class="stub"3n+4
n
,n=16,符合题意;
当m=3时,class="stub"19
9
=class="stub"3n+4
n
,n无正整数解;
当m=4时,class="stub"25
16
=class="stub"3n+4
n
,n无正整数解;
当m=5时,class="stub"31
25
=class="stub"3n+4
n
,n无正整数解;
当m=6时,class="stub"37
36
=class="stub"3n+4
n
,n无正整数解;
当m≥7时,m2-6m-1=(m-3)2-10>0,
class="stub"6m+1
m2
<1
,而class="stub"3n+4
n
=3+class="stub"4
n
>3

所以,此时不存在正整数m,n,且7<m<n,使得T1,Tm,Tn成等比数列.
综上,存在正整数m=2,n=16,且1<m<n,使得T1,Tm,Tn成等比数列.

更多内容推荐