a1=1,an,an+1为x^2-bn*x+t^n=0的根,0

Sn=a1+a2+...+an
bn=an+a(n+1)
b1+b2+...+bn
=a1+a2+a2+a3+...+an+a(n+1)
=Sn+S(n+1)-a1
=Sn+S(n+1)-1
lim(Sn+S(n+1)-1)≤3
limSn=limS(n+1)
所以
limSn≤2
ana(n+1)=t^n
a(n-1)an=t^(n-1)
相除得
a(n+1)/a(n-1)=t
即
a(2n+1)/a(2n-1)=t
a(2n)/a(2n-2)=t
a1=1 a2=t
{a(2n-1)} {a(2n)}均为等比为 t 的数列0