19. (12 points) Given the sequence $a_{n}$ with the first term $a_{1}=4$, the sum of the first $n$ terms is $S_{n}$, and
$$
S_{n+1}-3 S_{n}-2 n-4=0\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) Let the function
$$
f(x)=a_{n} x+a_{n-1} x^{2}+\cdots+a_{1} x^{n},
$$
$f^{\prime}(x)$ is the derivative of the function $f(x)$, and let $b_{n}=f^{\prime}(1)$, find the general term formula of the sequence $\left\{b_{n}\right\}$, and study its monotonicity.
19. (1) From $S_{n+1}-3 S_{n}-2 n-4=0$, we get
$$
S_{n}-3 S_{n-1}-2 n+2-4=0 \text {. }
$$

Subtracting the two equations, we get
$$
\begin{array}{l}
a_{n+1}-3 a_{n}-2=0 \\
\Rightarrow a_{n+1}+1=3\left(a_{n}+1\right)(n \geqslant 2) .
\end{array}
$$

Also, from the given, $a_{2}=14$, thus,
$$
a_{2}+1=3\left(a_{1}+1\right) \text {, }
$$

which means $\left\{a_{n}+1\right\}$ is a geometric sequence with the first term 5 and common ratio 3.
Therefore, $a_{n}=5 \times 3^{n-1}-1\left(n \in N_{+}\right)$.
(2) From $f^{\prime}(x)=a_{n}+2 a_{n-1} x+\cdots+n a_{1} x^{n-1}$,

we get
$$
\begin{aligned}
f^{\prime} & (1)=a_{n}+2 a_{n-1}+\cdots+n a_{1} \\
= & \left(5 \times 3^{n-1}-1\right)+2\left(5 \times 3^{n-2}-1\right)+ \\
& \cdots+n\left(5 \times 3^{0}-1\right) \\
= & 5\left[3^{n-1}+2 \times 3^{n-2}+\cdots+n \times 3^{0}\right]- \\
& \frac{n(n+1)}{2} .
\end{aligned}
$$

Let $S=3^{n-1}+2 \times 3^{n-2}+\cdots+n \times 3^{0}$.
Then $3 S=3^{n}+2 \times 3^{n-1}+\cdots+n \times 3^{1}$.
Subtracting, we get $S=-\frac{n}{2}-\frac{3-3^{n+1}}{4}$.
Thus, $f^{\prime}(1)=\frac{5 \times 3^{n+1}-15}{4}-\frac{n(n+6)}{4}$,
which means $b_{n}=\frac{5 \times 3^{n+1}-15}{4}-\frac{n(n+6)}{2}$.
And $b_{n+1}=\frac{5 \times 3^{n+2}-15}{4}-\frac{(n+1)(n+7)}{2}$.
Subtracting, we get $b_{n+1}-b_{n}=\frac{15 \times 3^{n}}{2}-n-\frac{7}{2}>0$.
Therefore, $b_{n}$ is a monotonically increasing sequence.