设函数 $f\left(x\right)=\frac{x}{x+2}(x>0)$,观察:
$f_1\left(x\right)=f\left(x\right)=\frac{x}{x+2}$

$f_2\left(x\right)=f\left(f_1\right(x\left)\right)=\frac{x}{3x+4}$

$f_3\left(x\right)=f\left(f_2\right(x\left)\right)=\frac{x}{7x+8}$

$f_4\left(x\right)=f\left(f_3\right(x\left)\right)=\frac{x}{15x+16}$

根据以上事实，由归纳推理可得：
当 $n\in N^{+}$且 $n\ge 2$时， $f_n\left(x\right)=f\left(f_{n-1}\right(x\left)\right)=$.