设sn是正项数列的前n项和且,sn=an的平方+二分之一an

a(n)>0.
s(n) = [a(n)]^2 + a(n)/2,
a(1) =s(1) = [a(1)]^2 + a(1)/2,
0 = [a(1)]^2 - a(1)/2 = a(1)[a(1)-1/2],a(1)=1/2.
s(n+1)=[a(n+1)]^2 + a(n+1)/2,
a(n+1)=s(n+1)-s(n) = [a(n+1)]^2 + a(n+1)/2 - [a(n)]^2 - a(n)/2,
0 = [a(n+1)]^2 - [a(n)]^2 - a(n+1)/2 - a(n)/2
= [a(n+1)+a(n)][a(n+1)-a(n)-1/2],
0 = a(n+1)- a(n)-1/2,
a(n+1) = a(n)+1/2,
{a(n)}是首项为a(1)=1/2,公差为1/2的等差数列.
a(n) = 1/2 + (n-1)/2 = n/2.