请在这里概述您的问题各项为正数的数列{an}的前n项和为Sn，且满足Sn=1/4an^2+1/2an+1/4(n∈N*)

(1) Sn=0.25an²+0.5an+0.25=0.25(an+1)² 即S(n-1)+an=0.25(an+1)² => S(n-1)=0.25(an-1)² 而S(n-1)=0.25(a(n-1)+1)² => (an-1)²=(a(n-1)+1)² ∵a1=s1=0.25(a1+1)² => a1=1 => (a2-1)²=4 =>a2=3 由此结合正项数列知an≥1,