已知各项均为正数的数列an的前n项和为Sn,满足8Sn=(an）^2+4an+3(n∈N+)且a1,a2,a7依次是等比

n=1时,
8S1=8a1=a1²+4a1+3
a1²-4a1+3=0
(a1-3)(a1-1)=0
a1=3或a1=1
n≥2时,
8Sn=an²+4an+3
8Sn-1=a(n-1)²+4a(n-1)+3
8Sn-8Sn-1=8an=an²+4an+3-a(n-1)²-4a(n-1)-3
an²-4an-a(n-1)²-4a(n-1)=0
[an²-a(n-1)²]-4[an+a(n-1)]=0
[an+a(n-1)][an-a(n-1)]-4[an+a(n-1)]=0
[an+a(n-1)][an-a(n-1)-4]=0
数列各项均为正,an+a(n-1)>0,要等式成立,则只有an-a(n-1)=4,为定值.
数列{an}是公差d为4的等差数列.
(1)
a1=1时,a2=a1+d=1+4=5 a7=a1+6d=1+24=25
a2²=25 a1×a7=1×25=25 a2²=a1×a7,a1、a2、a7成等比数列,满足题意.
(2)
a1=3时,a2=a1+d=3+4=7 a7=a1+6d=3+24=27
a2²=49,a1×a7=3×27=81 a2²≠a1×a7,a1、a2、a7不成等比数列,不满足题意,舍去.
综上,得数列{an}是以1为首项,4为公差的等差数列.
an=1+4(n-1)=4n-3
此时b1=a1=1 b2=a2=5 q=b2/b1=5,数列{bn}是以1为首项,5为公比的等比数列.
bn=1×5^(n-1)=5^(n-1)
数列{an}的通项公式为an=4n-3；数列{bn}的通项公式为bn=5^(n-1).