= Solution
Condition on the current generation size. By <independence> of offspring families, $\mathbb E[s^{Z_{n+1}}\mid Z_n=k]=G(s)^k$. Taking the <expected value> gives the <probability generating function> recursion
$$
G_{n+1}(s)=G_n(G(s)),\qquad G_0(s)=s.
$$
Induction therefore proves the <branching-process generating-function iteration>
$$
\boxed{G_n(s)=G^{\circ n}(s),}
$$
where the right side means $n$-fold composition, not an ordinary power.
Let $m=G'(1)$ and $\sigma^2=G''(1)+m-m^2$ be the offspring <mean> and <variance>, assuming a finite second moment. Differentiating the generating-function recursion at one gives $\mathbb EZ_n=m^n$ and, with $a_n=G_n''(1)$,
$$
a_{n+1}=m^2a_n+G''(1)m^n.
$$
Since $\operatorname{Var}(Z_n)=a_n+m^n-m^{2n}$, this implies
$$
v_{n+1}=m^2v_n+\sigma^2m^n,\qquad v_0=0.
$$
Iteration and summation of the <geometric progression> give the <mean and variance of a Galton-Watson generation>:
$$
\boxed{\operatorname{Var}(Z_n)=\begin{cases}
\displaystyle\sigma^2m^{n-1}\frac{m^n-1}{m-1},&m>0,\ m\ne1,\ n\ge1,\\
n\sigma^2,&m=1.
\end{cases}}
$$
For $m=0$ the offspring count is identically zero, so every later generation has zero <variance>. The zeroth generation always has zero <variance>. Finiteness of the needed moments is an assumption for this finite-valued <variance> formula, not a consequence of merely having a <probability generating function>. With finite positive <mean> but infinite offspring <variance>, all positive-generation variances are infinite.
For the total count, make the generation convention explicit. If the first $n$ generations include the ancestor, put $T_n=Z_0+\cdots+Z_{n-1}$ and $H_n(s)=\mathbb E[s^{T_n}]$, with $T_0=0$. Given $k$ children of the ancestor, each starts an independent depth-$n$ descendant tree with generating function $H_n$. The ancestor contributes an additional factor $s$, so
$$
\boxed{H_{n+1}(s)=sG(H_n(s)),\qquad H_0(s)=1,\quad H_1(s)=s.}
$$
This is the <finite-horizon total progeny generating function>. If instead “first $n$ generations” means descendants $Z_1+\cdots+Z_n$, excluding $Z_0$, write $\widehat H_n$ for that convention. Conditioning on the children now gives $\widehat H_{n+1}(s)=G(s\widehat H_n(s))$, with $\widehat H_0=1$ and $\widehat H_1=G$. The two conventions encode the same tree but different counts.
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