由题意得
(an +1)/2=√(Sn×1)
Sn=[(an +1)/2]²
n=1时,S1=a1=[(a1+1)/2]²,整理,得
(a1-1)²=0
a1=1
n≥2时,
Sn=[(an +1)/2]² S(n-1)=[(a(n-1) +1)/2]²
Sn-S(n-1)=an=[(an+1)/2]²-[(a(n-1)+1)/2]²
4an=an²+2an+1-[a(n-1)+1]²
(an -1)²-[a(n-1)+1]²=0
[an-1+a(n-1)+1][an -1-a(n-1)-1]=0
[an+a(n-1)][an-a(n-1)-2]=0
数列是正数数列,an+a(n-1)>0,要等式成立,只有an-a(n-1)=2,为定值.
数列{an}是以1为首项,2为公差的等差数列.
an=1+2(n-1)=2n-1
数列{an}的通项公式为an=2n-1.