10. (20 points) In the sequence $\left\{a_{n}\right\}$,
$$
\begin{array}{l}
a_{1}=1, \\
a_{n}=\frac{n}{n-1} a_{n-1}+2 n \times 3^{n-2}(n \in \mathbf{N}, n \geqslant 2) .
\end{array}
$$
(1) Find the general term $a_{n}$;
(2) Let $b_{n}=\frac{a_{n+1}}{n+1}\left(n \in \mathbf{Z}_{+}\right)$, prove: The sum of the first $n$ terms of the sequence $\left\{\frac{2 b_{n}}{\left(b_{n}-1\right)^{2}}\right\}$ is $S_{n}<2$.
10. (1) From the given condition, we have
$$
\begin{array}{l}
\frac{a_{n}}{n}-\frac{a_{n-1}}{n-1}=2 \times 3^{n-2} \\
\Rightarrow \frac{a_{n}}{n}=\sum_{i=2}^{n}\left(\frac{a_{i}}{i}-\frac{a_{i-1}}{i-1}\right)+a_{1} \\
\quad=2 \sum_{i=2}^{n} 3^{n-2}+1=3^{n-1} \\
\Rightarrow a_{n}=3^{n-1} n .
\end{array}
$$
(2) From the result of (1), we get $b_{n}=3^{n}$.
$$
\text { Then } \begin{array}{l} 
d_{n}=\frac{2 \times 3^{n}}{\left(3^{n}-1\right)^{2}} \\
= \frac{2 \times 3^{n-1}}{\left(3^{n}-1\right)\left(3^{n-1}-1\right)} \cdot \frac{3^{n}-3}{3^{n}-1} \\
<\frac{2 \times 3^{n-1}}{\left(3^{n}-1\right)\left(3^{n-1}-1\right)} \\
= \frac{1}{3^{n-1}-1}-\frac{1}{3^{n}-1} .
\end{array}
$$
$$
\begin{array}{l}
\text { Hence } S_{n}=\sum_{i=1}^{n} d_{i}<\frac{6}{4}+\sum_{i=2}^{n}\left(\frac{1}{3^{i-1}-1}-\frac{1}{3^{i}-1}\right) \\
=\frac{3}{2}+\frac{1}{2}-\frac{1}{3^{n}-1}<2 .
\end{array}
$$