6. Given $f(x)=\frac{1+x}{2-x}$, for $n \in \mathbf{N}$, define: $f_{1}(x)=f(x), f_{n+1}(x)=f\left[f_{n}(x)\right]$. If $f_{13}(x)=f_{31}(x)$, then the analytical expression of $f_{16}(x)$ is . $\qquad$
6. $\frac{x-1}{x}$.

It is easy to know that the inverse function of $f(x)$ is $f^{-1}(x)=\frac{2 x-1}{x+1}$. Then
$$
\begin{array}{l}
f^{-1}[f(x)]=\frac{2\left(\frac{1+x}{2-x}\right)-1}{\frac{1+x}{2-x}+1}=x . \\
\therefore f^{-1}\left[f_{n+1}(x)\right]=f^{-1}\left\{\left[f_{n}(x)\right]\right\}=f_{n}(x) . \\
\because f_{13}(x)=f_{31}(x), \\
\therefore f^{-1}\left[f_{13}(x)\right]=f^{-1}\left\{f\left[f_{31}(x)\right]\right\} . \\
\therefore f_{12}(x)=f_{30}(x) .
\end{array}
$$

Similarly, we have $f_{11}(x)=f_{29}(x), f_{10}(x)=$
$$
\begin{array}{l}
f_{28}(x), \cdots, f_{1}(x)=f_{19}(x) . \\
\therefore f^{-1}[f(x)]=f^{-1}\left[f_{19}(x)\right], \\
\therefore x=f_{18}(x) .
\end{array}
$$

Thus, $f^{-1}(x)=f_{17}(x)$,
$$
f^{-1}\left[f^{-1}(x)\right]=f_{16}(x) \text {. }
$$

Therefore, $f_{16}(x)=\frac{2\left(\frac{2 x-1}{x+1}\right)-1}{\frac{2 x-1}{x+1}+1}=\frac{x-1}{x}$.