数列各项均为正,Sn>0.2√Sn是a(n+2)与an的等比中项,则
(2√Sn)²=(an+2)an
4Sn=an²+2an
n=1时,4a1=4S1=a1²+2a1
a1²-2a1=0
a1(a1-2)=0
数列各项均为正,a1>0,因此只有a1-2=0
a1=2
n≥2时,Sn=(an²+2an)/4 S(n-1)=[a(n-1)²+2a(n-1)]/4
an=Sn-S(n-1)=(an²+2an)/4 -[a(n-1)²+2a(n-1)]/4
an²-a(n-1)²-2an-2a(n-1)=0
[an+a(n-1)][an-a(n-1)]-2[an+a(n-1)]=0
[an+a(n-1)][an-a(n-1)-2]=0
数列各项均为正,an+a(n-1)恒>0,因此只有an-a(n-1)-2=0
an-a(n-1)=2,为定值.数列{an}是以2为首项,2为公差的等差数列.
an=2+2(n-1)=2n