已知各项均为正数的数列 ｛an｝的前n项和满足Sn〉1,且6Sn=（an+1）（an+2）,n∈N* 求 （1)a1 （2）证明｛an｝是等差数列 并求通项公式

(1)
6Sn=(an+1)(an+2)
6S(n-1)=(a(n-1)+1)(a(n-1)+2)
两式想减得
6an=an^2+3an+2-a(n-1)^2-3a(n-1)-2
an^2-3an-a(n-1)^2-3a(n-1)=0
(an+a(n-1))(an-a(n-1))-3(an+a(n-1))=0
(an+a(n-1))(an-a(n-1)-3)=0
因为{an}各项均为正数，所以an+a(n-1)>0
所以an-a(n-1)-3=0
所以an-a(n-1)=3
所以{an}是等差数列
(2)
{an}公差为3，
6a1=(a1+1)(a1+2)=a1^2+3a1+2
a1^2-3a1+2=0
(a1-1)(a1-2)=0
a1=1或a1=2
因为Sn=a1>1,所以a1=2
所以an=2+3(n-1)=3n-1

(I)由a1=S1=1/6(a1+1)(a1+2),解得a1=1或a1=2,由假设a1=S1>1,因此a1=2,又由a(n+1)=S(n+1)-Sn=1/6(a(n+1)+1)(a(n+1)+2)-1/6(an+1)(an+2),得(a(n+1)+an)(a(n+1)-an-3)=0,即a(n+1)-an-3=0或a(n+1)=-an,因an>0,故a(n+1)=-...