Question

已知正项数列an,其前n项和为Sn,若6Sn=an^2+3an+2,a1,a3,a11成等比 求an的通项公式

Answer 1

如下：
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
a1,a3,a11成等比数列
a3^2=a1×a11
A2=A1+d;A4=A1+3d;A9=A1+8d
a3=a1+2d=a1+6
a11=a1+10d=a1+60
(a1+6)^2=a1×(a1+60)
a1=3/4 d=3
所以,通项式为：
an=3/4 +(n-1)×3
an=3n-9/4