In lecture today my professor introduced the notion of branching processes, and left the derivation of $\text{Var}(X_{n+1})$ to the students as an exercise. He said that we should try solving the problem through generating functions. For notation, let $\mu$ denote the average of the offspring distribution, and let $\sigma^2$ denote its variance. $X_n$ denotes the number of offspring present at time $n$, and it is assumed that $X_0=1$.
I began this by first deriving the recursive formula for $\text{Var}(X_{n+1})$, which goes as follows: $$\text{Var}(X_{n+1}) = \mu^2\text{Var}(X_n) + \mu^n\sigma^2$$ We can then define a generating function $f(x)$ as follows: $$\text{Var}(X_{n+1}) = c_{n+1}, f(x)=\sum_{n=0}^{\infty}c_nx^n=\sum_{n=0}^{\infty}c_{n+1}x^{n+1}$$ So we have: $$\sum_{n=0}^{\infty}c_{n+1}x^{n+1}=\sum_{n=0}^{\infty}(\mu^2 c_nx^{n+1})+\sum_{n=0}^{\infty}\mu^n\sigma^2x^{n+1}$$ I eventually simplify this down to the form of: $$f(x) = \sigma^2x\times\frac{1}{1-\mu x}\times\frac{1}{1-\mu^2x}$$ I'm new to generating functions, and don't have much experience with working with them. How do I pull out and isolate $c_{n+1}$ from the left-hand side of this expression? How would you guys finish this problem?
$$f(x) = \sigma^2 x \left(\sum_{n \ge 0} (\mu x)^n\right) \left(\sum_{m \ge 0} (\mu^2 x)^m\right)$$
In order to obtain $x^k$ from multiplying these terms, you get $1$ factor of $x$ from the first term $\sigma^2 x$, $n$ factors of $x$ from $(\mu x)^n$, and $m$ factors of $x$ from $(\mu^2 x)^m$. So collecting all cases where $1+n+m=k$, you have $$\sigma^2 \sum_{n=0}^{k-1} \mu^n \mu^{2(k-n-1)} = \sigma^2 \sum_{n=0}^{k-1} \mu^{2k-n-2}$$ as the coefficient for $x^k$.