题目

已知数列{an}是等比数列,其中a3=1,且a4,a5+1,a6成等差数列,数列{an/bn}的前n项和Sn=(n-1)2^(n-2)+1
（1）求数列{an}、{bn}的通项公式.
（2）设数列{bn}的前n项和为Tn,若T3n-Tn≥t对一切正整数n都成立,求实数t的取值范围.

提问时间：2020-08-03

答案

(1)
a4、a5+1、a6成等差数列,则2(a5+1)=a4+a6
a4=a3q a5=a3q² a6=a3q³ a3=1代入,整理,得
q³-2q²+q-2=0
q²(q-2)+(q-2)=0
(q²+1)(q-2)=0
q²+1恒为正,要等式成立,只有q=2
a1=a3/q²=1/2²=1/4
an=(1/4)×2^(n-1)=2^(n-3)
数列{an}的通项公式为an=2^(n-3).
S1=(1-1)×2^(1-2) +1=1 a1/b1=1 b1=a1=1/4
an/bn=Sn-Sn-1=(n-1)×2^(n-2)+1-(n-2)×2^(n-3)-1=n×2^(n-3)
bn=an/[n×2^(n-3)]=2^(n-3)/[n×2^(n-3)]=1/n
n=1时,b1=1/4,不满足.
数列{bn}的通项公式为
bn=1/4 n=1
1/n n≥2
[T3(n+3)-T(n+1)]-(T3n-Tn)
=[1/4+1/2+1/3+...+1/(3n)+1/(3n+1)+1/(3n+2)+1/(3n+3)]-[1/4+1/2+1/3+...+1/n+1/(n+1)]
-[1/4+1/2+1/3+...+1/(3n)]+(1+1/2+1/3+...+1/n)
=1/(3n+1)+1/(3n+2)+1/(3n+3)-1/(n+1)
>1/(3n+3)+1/(3n+3)+1/(3n+3)-1/(n+1)
=1/(n+1)-1/(n+1)=0
T3(n+3)-T(n+1)>T3n-Tn
即随n增大,T3n-Tn单调递增,当n=1时,T3n-Tn取得最小值.
T3-T1=(1/4+1/2+1/3)-(1/4)=1/2+1/3=5/6
要不等式T3n-Tn≥t对于一切正整数n恒成立,只要t≤5/6.