9. 若数列 $\left\{a_{n}\right\}$ 满足: $a_{1}=1$, 且 $\frac{a_{n+1}}{a_{n}}=\frac{n+2}{n}\left(n \in \mathbf{N}^{*}\right)$, 则当 $n \geqslant 2$ 时, $a_{n}=$
$\because a_{1}=1$, 且 $\frac{a_{n+1}}{a_{n}}=\frac{n+2}{n}\left(n \in \mathbf{N}^{*}\right)$.

$\therefore \frac{a_{2}}{a_{1}} \cdot \frac{a_{3}}{a_{2}} \cdot \frac{a_{4}}{a_{3}} \cdots \cdot \frac{a_{n-1}}{a_{n-2}} \cdot \frac{a_{n}}{a_{n-1}}$

$=\frac{3}{1} \cdot \frac{4}{2} \cdot \frac{5}{3} \cdots \frac{n}{n-2} \cdot \frac{n+1}{n-1}$,

即 $a_{n}=\frac{n(n+1)}{2}$.