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,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.
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答案:
分析:(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,由此能求出an和Sn.
(2)由bn=anan+1=(3n-2)(3n+1),知
=
=
(
-
),由此能够证明Tn<
.
(3)由(2)知,Tn=
,故T1=
,Tm=
,Tn=
,由T1,Tm,Tn成等比数列,能够推导出存在正整数m=2,n=16,且1<m<n,使得T1,Tm,Tn成等比数列.