由已知an與1的等差中項(xiàng)等于Sn與1的等比中項(xiàng)得
(an +1)/2=√Sn
Sn=(an +1)²/4
n=1時(shí),S1=a1=(a1+1)²/4,整理,得
(a1-1)²=0
a1=1
n≥2時(shí),
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),數(shù)列各項(xiàng)均為正,舍去)
an=a(n-1)+2
數(shù)列{an}是以1為首項(xiàng),2為公差的等差數(shù)列.
an=1+2(n-1)=2n-1
a1=1 a2=2×2-1=3 a3=2×3-1=5