1. Find the value of the expression $(\ldots(((2 * 3) * 4) * 5) * \ldots) * 1995$, where $x * y=\frac{x+y}{1+x y}$.

(Macedonia)
1. Note that

$$
\frac{1-x * y}{1+x * y}=\frac{1-x}{1+x} \cdot \frac{1-y}{1+y}
$$

If the value of the desired expression is $P$, it follows that $\frac{1-P}{1+P}=\frac{1-2}{1+2} \cdot \frac{1-3}{1+3} \cdots \frac{1-1995}{1+1995}=$ $\frac{2 \cdot 1994!}{1996!}=\frac{1}{998 \cdot 1995}$, hence $P=\frac{998 \cdot 1995-1}{998 \cdot 1995+1}$.

Second solution. By induction, it can be shown that $f(n)=(\ldots((2 * 3) * 4) * \ldots) * n=\frac{n(n+1)-2}{n(n+1)+2}$. Indeed, this is true for $n=2$, and for $n>2$ we have $f(n)=f(n-1) * n=\frac{n(n-1)-2}{n(n-1)+2} * n=\frac{n(n+1)-2}{n(n+1)+2}$.