若数列An是有界数列,A(n+1)=-An^2+2tAn,A1=t(t>0),求实数t的取值范围.

Question

若数列An是有界数列,A(n+1)=-An^2+2tAn,A1=t(t>0),求实数t的取值范围.

Answer 1

A(n+1)=-An^2+2tAn
则A(n+1)-1=-An^2+2tAn-1=-（An -1）^2
于是An -1=-[A(n-1) -1]^2
=-{-[A(n-2) -1]^2}^2=-[A(n-2) -1]^(2^2)
=.
=-(A1 -1)^[2^(n-1)]
=-(t-1)^[2^(n-1)]
若数列An是有界数列,则An -1也是有界数列;
则lim(n→+∞) An -1=lim(n→+∞) -(t-1)^[2^(n-1)]
=-lim(n→+∞) (t-1)^[2^(n-1)]是有限值.
那么就要求-1≤t-1≤1.
∴0≤t≤2
加分!