fengguobaiyu
幼苗
共回答了10个问题采纳率:100% 举报
(Ⅰ)设数列{a n }的公差为d,由a 3 =a 1 +2d=7,a 1 +a 2 +a 3 =3a 1 +3d=12.
解得a 1 =1,d=3∴a n =3n-2
∵f(x)=x 3 ∴S n = f(
3 a n+1
) =a n+1 =3n+1.
(Ⅱ)b n =a n S n =(3n-2)(3n+1)
∴
1
b n =
1
(3n-2)(3n+1) =
1
3 (
1
3n-2 -
1
3n+1 ) ∴ T n =
1
3 (1-
1
3n+1 )<
1
3
(Ⅲ)由(2)知, T n =
n
3n+1 ∴ T 1 =
1
4 , T m =
m
3m+1 , T n =
n
3n+1 ∵T 1 ,T m ,T n 成等比数列.
∴ (
m
3m+1 ) 2 =
1
4
n
3n+1 即
6m+1
m 2 =
3n+4
n
当m=1时,7=
3n+4
n ,n=1,不合题意;当m=2时,
13
4 =
3n+4
n ,n=16,符合题意;
当m=3时,
19
9 =
3n+4
n ,n无正整数解;当m=4时,
25
16 =
3n+4
n ,n无正整数解;
当m=5时,
31
25 =
3n+4
n ,n无正整数解;当m=6时,
37
36 =
3n+4
n ,n无正整数解;
当m≥7时,m 2 -6m-1=(m-3) 2 -10>0,则
6m+1
m 2 <1 ,而
3n+4
n =3+
4
n >3 ,
所以,此时不存在正整数m,n,且1<m<n,使得T 1 ,T m ,T n 成等比数列.
综上,存在正整数m=2,n=16,且1<m<n,使得T 1 ,T m ,T n 成等比数列.
1年前
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