28. A sequence $\left\{a_{n}\right\}$ is defined by $\mathrm{a}_{1}-2, a_{n}=\frac{1+a_{n-1}}{1-a_{n-1}}, \mathrm{n} \geq 2$. Find the value of $-2008 a_{2007}$.
28. Answer: 1004
Since $a_{1}=2, a_{2}=-3, a_{3}=-\frac{1}{2}, a_{4}=\frac{1}{3}, a_{5}=2$, thus we consider
$$
\begin{array}{l}
a_{n+4}=\frac{1+a_{n+3}}{1-a_{v+3}}=\frac{1+\frac{1+a_{n+2}}{1-a_{n+2}}}{1-\frac{1+a_{n+2}}{1-a_{n+2}}}=-\frac{1}{a_{n+2}}=-\left(\frac{1-a_{n+1}}{1+a_{n+1}}\right) \\
=-\frac{1-\frac{1+a_{n}}{1-a_{n}}}{1+\frac{1+a_{n}}{1-a_{n}}}=a_{n} \\
a_{2007}=a_{4 \times 501+3}=a_{3}=-\frac{1}{2} \\
-2008 a_{2007}=1004 \\
\end{array}
$$