设正数列{an}的前n项和为Sn,且根号下Sn是an和1的等差中项,

2*Sn^(1/2)=An+1
(1) 2*S1^(1/2)=A1+1,S1=A1
A1=1
(2)
Sn=(An+1)^2/4
S(n-1)=[A(n-1)+1]^2/4
An=Sn-S(n-1)=(1/4)*((An+1)^2-[A(n-1)+1]^2)
整理得到An-A(n-1)=2
等差数列
(3)
An=A1+(n-1)*d=1+(n-1)*2=2n-1