100 Given $f(x)=\frac{1+x}{2-x}$, for $n \in \mathbf{N}$, define $f_{1}(x)=f(x), f_{n+1}(x)=f\left[f_{n}(x)\right]$, assuming $f_{13}(x)=f_{31}(x)$, then the analytical expression for $f_{16}(x)$ is
A. $\frac{x+1}{x}$
B. $\frac{x}{x+1}$
C. $\frac{x-1}{x}$
D. $\frac{x}{x-1}$
100 C. The inverse function of $f(x)$ is
$$
f^{-1}(x)=\frac{2 x-1}{x+1} .
$$

Since $f_{13}(x)=f_{31}(x)$, we have
$$
f^{-1}\left[f_{13}(x)\right]=f^{-1}\left[f_{31}(x)\right]
$$

which means $f_{12}(x)=f_{30}(x)$. Similarly, we can get
$$
\begin{array}{c}
f_{11}(x)=f_{29}(x), \cdots, f(x)=f_{19}(x), \\
x=f_{18}(x), f^{-1}(x)=f_{17}(x), \\
f_{16}(x)=f^{-1}\left(f^{-1}\left(x_{1}\right)\right)=\frac{2 f^{-1}(x)-1}{f^{-1}(x)+1}=\frac{2\left(\frac{2 x-1}{x+1}\right)-1}{\frac{2 x-1}{x+1}+1}=\frac{3 x-3}{3 x}=\frac{x-1}{x} .
\end{array}
$$