20. Let $a_{1}, a_{2}, \ldots$ be a sequence of rational numbers such that $a_{1}=2$ and for $n \geq 1$
$$
a_{n+1}=\frac{1+a_{n}}{1-a_{n}} .
$$

Determine $30 \times a_{2008}$.
20. Answer: 10 .
We have
$$
\begin{array}{l}
a_{n+1}=\frac{1+a_{n}}{1-a_{n}} \\
a_{n+2}=\frac{1+a_{n+1}}{1-a_{n+1}}-\frac{1+\frac{1+\varepsilon_{n}}{1-\varepsilon_{n}}}{1-\frac{1+\varepsilon_{n}}{1-\varepsilon_{n}}}=-\frac{1}{a_{n}} \\
a_{n+4}=-\frac{1}{a_{n+2}}=a_{n} .
\end{array}
$$

Thus $a_{2100 s}=a_{2004}=\cdots=a_{4}=\frac{1}{3}$