9. For the sequence $\left\{a_{n}\right\}$, the adjacent terms $a_{n}$ and $a_{n+1}$ are the two roots of the quadratic equation $x^{2}-3 n x+c_{n}=0(n=1,2,3, \cdots)$. When $a_{1}=1$, find $c_{n}$.

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$$
\begin{array}{l}
a_{n}+a_{n+1}=3 n \Rightarrow a_{n+1}=-a_{n}+3 n, \text{ let } a_{n+1}+k(n+1)+b=-\left(a_{n}+k n+b\right) \\
\Rightarrow a_{n+1}=-a_{n}-2 k n-k-2 b \Rightarrow k=-\frac{3}{2}, b=\frac{3}{4} . \\
\text{ Thus } a_{n}-\frac{3}{2} n+\frac{3}{4}=(-1)^{n-1}\left(a_{1}-\frac{3}{2}+\frac{3}{4}\right)=(-1)^{n-1} \cdot \frac{1}{4} \\
\Rightarrow a_{n}=\frac{3}{2} n-\frac{3}{4}+(-1)^{n-1} \cdot \frac{1}{4}=\frac{3}{2} n-\frac{1}{4}\left[3+(-1)^{n}\right] . \\
\text{ That is }\left\{\begin{array} { l } 
{ a _ { 2 n - 1 } = 3 n - 2 , } \\
{ a _ { 2 n } = 3 n - 1 . }
\end{array} \text{ Therefore } \left\{\begin{array}{l}
c_{2 n-1}=a_{2 n-1} a_{2 n}=9 n^{2}-9 n+2, \\
c_{2 n}=a_{2 n} a_{2 n+1}=9 n^{2}-1 .
\end{array}\left(n \in \mathbf{N}^{*}\right) .\right.\right.
\end{array}
$$