22. Given that $a_{n+1}=\frac{a_{n-1}}{1+n a_{n-1} a_{n}}$, where $n=1,2,3, \ldots$, and $a_{0}=a_{1}=1$, find the value of $\frac{1}{a_{190} a_{200}}$.
22. Answer: 19901
First note that $\frac{1}{a_{n+1} a_{n}}-\frac{1}{a_{n} a_{n-1}}=\frac{1+n a_{n-1} a_{n}}{a_{n-1} a_{n}}-\frac{1}{a_{n} a_{n-1}}=n$. Therefore
$$
\begin{array}{l}
\qquad \sum_{n=1}^{109}\left(\frac{1}{a_{n+1} a_{n}}-\frac{1}{a_{n} a_{n-1}}\right)=\sum_{n=1}^{199} n=\frac{199 \times 200}{2}=19900 . \\
\text { Hence } \frac{1}{a_{200} a_{199}}=1+19900=19901 .
\end{array}
$$