数列{An},An>0,前n项和为Sn,A1=2 An=2倍根号下（2S（n-1））再加上2 求An的通项数列{An},An>0,前n项和为Sn,A1=2 An=2倍根号下（2S（n-1））再加上2 求An的通项

A1=2
A2=2*根号(2*S1)+2=2*根号(2*A1)+2=6
n>=2时
An=2*根号(2*S(n-1))+2
(An-2)^2=8*S(n-1)
An^2-4*An+4=8*S(n-1)
n>=3时
A(n-1)^2-4*A(n-1)+4=8*S(n-2)
两式相减
An^2-A(n-1)^2-4*(A(n)-A(n-1))=8*A(n-1)
(An+A(n-1))*(An-A(n-1)-4)=0
An>0
则An-A(n-1)-4=0
A1=2
则An=4n-2

an=2√2Sn-1+2
(an-2)^2=8Sn-1
[a(n+1)-2]^2=8Sn
8Sn-8Sn-1=8an=[a(n+1)-2]^2-(an-2)^2=[an+a(n+1)-4][a(n+1)-an]
-an^2+a(n+1)^2-4a(n+1)-4an=0
a(n+1)^2-4a(n+1)+4-an^2-4an-4=0
[a(n+1)-2]^2=(an+2)^2
a(n+1)=an+4或a(n+1)=-an(an恒>0,舍去)
an=a(n-1)+4,数列为首项是2,公差是4的等差数列.
an=2+4(n-1)=4n-2
数列{an}的通项公式为an=4n-2