16. Let $\left\{a_{n}\right\}$ be a sequence of positive integers such that $a_{1}=1, a_{2}=2009$ and for $n \geq 1$, $a_{n+2} a_{n}-a_{n+1}^{2}-a_{n+1} a_{n}=0$. Determine the value of $\frac{a_{993}}{100 a_{991}}$.
16. Answer: 89970
$$
a_{n+2} a_{n}-a_{n+1}^{2}-a_{n+1} a_{n}=0 \quad \Rightarrow \quad \frac{a_{n+2} a_{n}-a_{n+1}^{2}-a_{n+1} a_{n}}{a_{n+1} a_{n}}=0
$$
$\frac{a_{n+2}}{a_{n+1}}-\frac{a_{n+1}}{a_{n}}=1$. From here, we see that $\left\{\frac{a_{n+1}}{a_{n}}\right\}$ is an arithmetic sequence with first term 2009 and common difference 1 . Thus $\frac{a_{n+1}}{a_{n}}=n+2008$, and that $\frac{a_{993}}{a_{992}}=992+2008=3000$ and $\frac{a_{992}}{a_{991}}=991+2008=2999$. We therefore have $\frac{a_{993}}{100 a_{991}}=30(2999)=89970$