数列问题数列问题

6*Sn=An^2+3*An+2
6Sn=(An+1)(An+2),
6(a1+a2+a3+...+an)=(an+1)(an+2),
6(a1+a2+a3+...+an-1)=a²n-3an+2=(an-1)(an-2),
6(a1+a2+a3+...+an-1)=6Sn-1
6S（n-1）=(a(n-1)+1)(a（n-1）+2)
(a(n-1)+1)(a（n-1）+2)=(an-1)(an-2),
a²(n-1)+3a(n-1)=a²n-3an
a²n-a²(n-1)=3an+3a(n-1)
（an-an-1)(an+an-1)=3(an+an-1)
an-an-1=3.
所以an为等差数列
s1=a1=1/6(A1+1)(A1+2),
解得a1=1或a1=2
A2,A4,A9成等比数列
所以验证得a1=1
所以an=3n-2

6*Sn=An^2+3*An+2
6*S(n+1)=A(n+1)^2+3*A(n+1)+2
两式相减,得
6*A(n+1)=A(n+1)^2+3*A(n+1)-An^2-3*An
整理得
[A(n+1)-An-3]×[A(n+1)+An]=0
因为An为正数
所以
A(n+1)=An+3
A2=A1+3
A4=A1+9
A9=A1+24
所以
由A2×A9=A4^2解得
A1=1
所以An=3n-2