设sn是正项数列的前n项和且,sn=an的平方+二分之一an
问题描述：
设sn是正项数列的前n项和且,sn=an的平方+二分之一an
答
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)...