(14) Let the sequence $\left\{a_{n}\right\}\left(a_{n} \geqslant 0\right)$ satisfy $a_{1}=0, a_{2}=1, a_{3}=9$, and
$$
S_{n}^{2} S_{n-2}=10 S_{n-1}^{3}(n>3),
$$

where $S_{n}$ is the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$. Find the expression for $a_{n}(n \geqslant 3)$.
(14) Because $S_{n}^{2} S_{n-2}=10 S_{n-1}^{3}$, so
$$
\left(\frac{S_{n}}{S_{n-1}}\right)^{2}=10 \frac{S_{n-1}}{S_{n-2}} .
$$

Let $b_{n}=\frac{S_{n}}{S_{n-1}}(n \geqslant 3)$, then $b_{n}=\left(10 b_{n-1}\right)^{\frac{1}{2}}$, therefore
$$
\begin{array}{l}
b_{n}=10^{\frac{1}{2}}\left(10^{\frac{1}{2}} b_{n-2}^{\frac{1}{2}}\right)^{\frac{1}{2}} \\
=10^{\frac{1}{2}+\frac{1}{4}} b_{n-2}^{\frac{1}{n}} \\
\text {... } \\
=10^{\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^{n-3}}} b^{\frac{1}{2^{n-3}}} \\
=10 \text {, } \\
\end{array}
$$

Then $S_{n}=10 S_{n-1}$, and
$$
a_{n}=9 S_{n-1}=9 a_{n-1}+9 S_{n-2}=10 a_{n-1},
$$

and $a_{3}=9$, so, $a_{n}=9 \times 10^{n-3}(n \geqslant 3)$.