Sn=(an+1)^2/4=(an^2+2an+1)/4
Sn-1=[a(n-1)+1]^2=[(a(n-1)^2+2a(n-1)+1]/4
Sn-Sn-1=an=[an^2+2an-a(n-1)^2-2a(n-1)]/4
4an=an^2+2an-a(n-1)^2-2a(n-1)
an^2-2an=a(n-1)^2+2a(n-1)
(an-1)^2=[a(n-1)+1]^2
又a1=1,an>0
an-1=a(n-1)+1
an-a(n-1)=2
数列为等差数列,首项为1,公差为2.
an=1+(n-1)*2=2n+1