= Solution
With population $n$, the transition diagram has the two outgoing arrows
$$
n\xrightarrow{\lambda}n+3,
\qquad
n\xrightarrow{\beta n}n-1.
$$
Thus this is a <batch-birth linear-death process>. If $P_n(t)=\mathbb P(N_t=n)$ and $P_j=0$ for $j<0$, probability enters state $n$ from $n-3$ by a birth and from $n+1$ by a death. Hence the master equation is
$$
\boxed{
\dot P_n
=\lambda P_{n-3}+\beta(n+1)P_{n+1}
-(\lambda+\beta n)P_n
}.
$$
The <Markov jump-process generator> is
$$
(Lf)(n)=\lambda\bigl(f(n+3)-f(n)\bigr)
+\beta n\bigl(f(n-1)-f(n)\bigr).
$$
For $f(n)=n$, this gives $Ln=3\lambda-\beta n$. Therefore the <expected value> $m(t)=\mathbb E[N_t]$ satisfies
$$
m'=3\lambda-\beta m,
\qquad m(0)=n_0,
$$
so
$$
\boxed{
m(t)=\frac{3\lambda}{\beta}(1-e^{-\beta t})+n_0e^{-\beta t}
}.
$$
For $f(n)=n^2$,
$$
Ln^2
=\lambda(6n+9)+\beta n(-2n+1).
$$
Writing $s=\mathbb E[N_t^2]$ and using the generator identity gives
$$
s'=6\lambda m+9\lambda-2\beta s+\beta m.
$$
The <variance> $v=s-m^2$ consequently obeys
$$
\begin{aligned}
v'&=s'-2mm'\\
&=9\lambda+\beta m-2\beta v.
\end{aligned}
$$
If $v(0)=v_0<\infty$, substitution of the formula for $m$ yields
$$
v(t)=v_0e^{-2\beta t}
+\frac{6\lambda}{\beta}(1-e^{-2\beta t})
+\left(n_0-\frac{3\lambda}{\beta}\right)
(e^{-\beta t}-e^{-2\beta t}).
$$
All transient terms vanish, and the <moments of a batch-birth linear-death process> therefore give
$$
\boxed{\operatorname{var}(N_t)\longrightarrow\frac{6\lambda}{\beta}}.
$$
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