因为an,Sn,an^2成等差数列
所以2Sn=an^2+an
2an=2Sn-2S(n-1)
=an^2+an-a(n-1)^2-a(n-1)
得:(an-a(n-1))(an+a(n-1))-(an+a(n-1))=0
(an+a(n-1))(an-a(n-1)-1)=0
因为数列{an}为正项数列
所以an-a(n-1)=1
即{an}是公差为1的等比数列
2a1=a1^2+a1 得:a1=1
所以an=1+(n-1)*1=n