3. Let the function $f(x)=\frac{1}{2}+\log _{2} \frac{x}{1-x}$, define $S_{n}$ $=\sum_{i=1}^{n-1} f\left(\frac{i}{n}\right)$, where $n \in \mathbf{N}^{*}$, and $n \geqslant 2$, then $S_{n}=$
3. $\frac{n-1}{2}$.

Solution: When $x_{1}+x_{2}=1$, $f\left(x_{1}+f\left(x_{2}\right)\right)=1+$
$$
\begin{array}{l}
\log _{2} \frac{x_{1} x_{2}}{\left(1-x_{1}\right)\left(1-x_{2}\right)}=1, \\
2 S_{n}=\sum_{i=1}^{n-1}\left[f\left(\frac{i}{n}\right)+f\left(\frac{n-i}{n}\right)\right]=n-1, \text{ hence } S_{n} \\
=\frac{n-1}{2} .
\end{array}
$$