题目

数列{F(n)}的递推公式为：F(n+1)F(n-1)=F(n)^2+1,前两项为：F(1)=1,F(2)=2.求通项公式.

提问时间：2020-11-16

答案

Fn+1Fn-1=Fn^2+1Fn+2Fn=Fn+1^2+1两式相减得：Fn+2Fn-Fn+1Fn-1=Fn+1^2-Fn^2移项得：Fn+2Fn+Fn^2=Fn+1Fn-1+Fn+1^2即Fn(Fn+2+Fn)=Fn+1(Fn+1+Fn-1)则(Fn+2+Fn)/Fn+1=(Fn+1+Fn-1)/Fn即数列(Fn+2+Fn)/Fn+1为常数列可得(Fn+...