= Solution
Summing the geometric series for the offspring <probability generating function> gives
$$
F_1(s)=\sum_{k=0}^\infty\frac{s^k}{2^{k+1}}=\frac1{2-s}.
$$
The fixed-point equation $s=1/(2-s)$ is $(s-1)^2=0$, whose only root in $[0,1]$ is one. The <Galton-Watson extinction fixed point> result therefore proves \b[extinction is certain].
For the explicit iterates, the proposed formula at $n=1$ is $1/(2-s)$, so the base case holds. If it holds at $n$, part (a) gives
$$
F_{n+1}(s)=
\frac{n-(n-1)/(2-s)}{n+1-n/(2-s)}
=\frac{(n+1)-ns}{(n+2)-(n+1)s}.
$$
This is exactly the same formula with $n$ replaced by $n+1$. By induction,
$$
\boxed{F_n(s)=\frac{n-(n-1)s}{(n+1)-ns},\qquad n\geq1.}
$$
Evaluating the <probability generating function> at zero selects the mass of an empty generation:
$$
\boxed{P(X_n=0)=F_n(0)=\frac n{n+1}.}
$$
These <probabilities> increase to one, also confirming the extinction conclusion directly.
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