由已知an与1的等差中项等于Sn与1的等比中项得
(an +1)/2=√Sn
Sn=(an +1)²/4
n=1时,S1=a1=(a1+1)²/4,整理,得
(a1-1)²=0
a1=1
n≥2时,
Sn=(an+1)²/4 Sn-1=[a(n-1)+1]²/4
Sn-Sn-1=an=(an+1)²/4 -[a(n-1)+1]²/4
(an-1)²=[a(n-1)+1]²
an -1=a(n-1)+1或an -1=-a(n-1)-1(an=-a(n-1),数列各项均为正,舍去)
an=a(n-1)+2
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
a1=1 a2=2×2-1=3 a3=2×3-1=5