(9) (16 points) Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=3, a_{n}+a_{n-1}=2$ $\frac{n(3 n+1)}{a_{n}-a_{n-1}}\left(n \geqslant 2, a_{n}>0\right)$. Find the general term formula of the sequence $\left\{a_{n}\right\}$.
9
$$
\begin{array}{l}
a_{n}^{2}-a_{n-1}^{2}=2\left(a_{n}-a_{n-1}\right)+n(3 n+1) \\
\Rightarrow a_{n}^{2}-2 a_{n}=a_{n-1}^{2}-2 a_{n-1}+3 n^{2}+n, \\
a_{n-1}^{2}-2 a_{n-1}=a_{n-2}^{2}-2 a_{n-2}+3(n-1)^{2}+(n-1), \\
\cdots \cdots \cdots \\
a_{2}^{2}-2 a_{2}=a_{1}^{2}-2 a_{1}+3 \times 2^{2}+2 .
\end{array}
$$

Adding the above $n-1$ equations, we get
$$
\begin{array}{l}
a_{n}^{2}-2 a_{n}=3^{2}-2 \times 3+3 \times\left[\frac{1}{6} n(n+1)(2 n+1)-1\right]+ \\
\frac{1}{2}(n-1)(n+2) \\
\Rightarrow\left(a_{n}-1\right)^{2}=n^{3}+2 n^{2}+n .
\end{array}
$$

Since $a_{n}>0$, we have $a_{n}=1+(n+1) \sqrt{n}\left(n \in \mathbf{N}^{*}\right)$.