(1)
a4、a5+1、a6成等差數(shù)列,則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恒為正,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)
數(shù)列{an}的通項(xiàng)公式為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時(shí),b1=1/4,不滿足.
數(shù)列{bn}的通項(xiàng)公式為
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
即隨n增大,T3n-Tn單調(diào)遞增,當(dāng)n=1時(shí),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對(duì)于一切正整數(shù)n恒成立,只要t≤5/6.