已知数列{An}的各项均为正数,前n项和Sn满足6Sn=An^2+3An+2,若A2,A4,A9成等比数列,则数列{an}的通项an=?

题目

已知数列{An}的各项均为正数,前n项和Sn满足6Sn=An^2+3An+2,若A2,A4,A9成等比数列,则数列{an}的通项an=?

答案

6Sn=An^2+3An+2
6S(n-1)=[A(n-1)]^2+3A(n-1)+2
6Sn-6S(n-1)=6An=An^2+3An+2-{[A(n-1)]^2+3A(n-1)+2}
An-A(n-1)=3
{An}为等差数列,d=3
A4^2=A2A9
A2=A1+d;A4=A1+3d;A9=A1+8d
分别代入得
A1=d/3=1
An=1+3(n-1)=3n-2