14. Let $\left\{a_{n}\right\}$ be a sequence of positive numbers, with the sum of its first $n$ terms being $S_{n}$, and for all natural numbers $n$, we have
$$
S_{n}=\frac{3 n+1}{2}-\frac{n}{2} a_{n} \text {. }
$$
(1) Write down the first three terms of the sequence $\left\{a_{n}\right\}$;
(2) Find the general term formula for the sequence $\left\{a_{n}\right\}$ (include the derivation process);
(3) Let $b_{n}=\frac{1}{a_{n}-1}\left(n \in \mathbf{N}_{+}\right)$, find the sum of the first $n$ terms of the sequence $\left\{b_{n}\right\}$.
14. (1) According to the problem, $S_{n}=\frac{3 n+1}{2}-\frac{n}{2} a_{n}\left(a_{n}>0\right)$.

When $n=1$, $S_{1}=\frac{3+1}{2}-\frac{1}{2} a_{1}=a_{1}$, solving gives $a_{1}=\frac{4}{3}$.

When $n=2$, $S_{2}=\frac{6+1}{2}-\frac{2}{2} a_{2}=a_{1}+a_{2}$, solving gives $a_{2}=\frac{13}{12}$.

When $n=3$, $S_{3}=\frac{9+1}{2}-\frac{3}{2} a_{3}=a_{1}+a_{2}+a_{3}$, solving gives $a_{3}=\frac{31}{30}$.
Thus, the first three terms of the sequence are $\frac{4}{3}, \frac{13}{12}, \frac{31}{30}$.
(2) According to the problem,
$$
S_{n}=\frac{3 n+1}{2}-\frac{n}{2} a_{n}\left(n \in \mathbf{N}_{+}\right) \text {. }
$$

From this, we get $S_{n+1}=\frac{3 n+4}{2}-\frac{n+1}{2} a_{n+1}$.
Therefore, $a_{n+1}=S_{n+1}-S_{n}$
$$
=\frac{3}{2}+\frac{n}{2} a_{n}-\frac{n+1}{2} a_{n+1} \text {. }
$$

Rearranging gives $\frac{(n+3)}{2}\left(a_{n+1}-1\right)=\frac{n}{2}\left(a_{n}-1\right)$.
Thus, $\frac{(n+3)(n+2)(n+1)}{2}\left(a_{n+1}-1\right)$
$=\frac{(n+2)(n+1) n}{2}\left(a_{n}-1\right)$
$$
\begin{array}{l}
=\cdots \\
=\frac{(1+2)(1+1) \times 1}{2}\left(a_{1}-1\right)=1 .
\end{array}
$$

Then, $a_{n}=1+\frac{2}{n(n+1)(n+2)}$.
$$
\begin{array}{l}
\text { (3) Clearly, } b_{n}=\frac{n(n+1)(n+2)}{2} \\
=\frac{n(n+1)(n+2)(n+3)}{8}- \\
\frac{(n-1) n(n+1)(n+2)}{8} .
\end{array}
$$

Thus, $S_{n}^{\prime}=\frac{n(n+1)(n+2)(n+3)}{8}$.