S(n+1)=(a(n+1)+2)^2/8 a(n+1)=S(n+1)-S(n) 化简可得（a(n+1)）^2-(a(n))^2=4*(a(n)+a(n+1)) 所以a(n+1)-a(n)=4 故得证,3,已知数列{an}中,an>0其前n项和为Sn,且Sn=1/8(an+2)²,求证：数列{an}为等差数列
