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}為等差數(shù)列,d=3
a1,a3,a11成等比數(shù)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,通項式為：
an=3/4 +(n-1)×3
an=3n-9/4額答案為3n-1你好，不好意思，我算a1那步60那里應該是30.(馬虎了)6Sn=an^2+3an+26S(n-1)=[a(n-1)]^2+3a(n-1)+26Sn-6S(n-1)=6an=an^2+3an+2-{[a(n-1)]^2+3a(n-1)+2}an-a(n-1)=3{an}為等差數(shù)列，d=3a1,a3,a11成等比數(shù)列a3^2=a1×a11a3=a1+2d=a1+6a11=a1+10d=a1+30(a1+6)^2=a1×(a1+30)a1=2 d=3所以，通項式為：an=2 +(n-1)×3an=3n-1如上，這次對了。嘎嘎嘎，不好意思O(∩_∩)O~