已知数列{an}中,an>0,Sn=a1+a2+……+an,且an=6Sn/（an+3) 求Sn

an=6Sn/（an+3)
an^2+3an=6Sn 1)
a(n-1)^2+3a(n-1)=6S(n-1) 2)
an=Sn-S(n-1) 3)
1)-2)再代入3)整理有
(An-A(n-1))*(An+A(n-1))=3*(An+A(n-1)) 4)
An>0
An+A(n-1)>0
由4)得到
An-A(n-1)=3
3为常数,所以An为等差数列
A1^2+3*A1=6S1=6A1
A1=3
An=A1+(n-1)*d=3+(n-1)*3=3n
Sn=A1+A2+...+An=3*(1+2+3+...+n)=3*(1+n)*n/2

由题意知：an=6Sn/（an+3)
an^2+3an=6Sn
a(n-1)^2+3a(n-1)=6S(n-1)
6an=6Sn-6S(n-1)
(an-a(n-1))*(an+a(n-1))=3*(an+a(n-1))
an>0
an+a(n-1)>0
因此
an-...