I'm having a problem understanding a part of an article and I would appreciate help.
The author is trying to derive recurrence relations from generating functions of the form
$$\frac{P_{dn}(t)}{(1-t)^{dn+1}}=\sum_{k\geq 0} f_n(k)t^k.$$ We note that $f_n$ has degree $d(n+1)$ if and only if $P(1)\neq0.$
The argument that he uses start with the following idea:
Suppose $f_{n+1}(k)=g_{n}(k)f_n{(k)}$ for some degree $d$ polynomial $g_n(k)$ for all $n$. For instance let $d=1$. In general, if $g_n(k)=\alpha_nk+\beta_n$ for some $\alpha_n,\beta_n\in\mathbb{R}$, we have
\begin{equation} \frac{P_n(t)}{(1-t)^{n+1}}=\alpha_nt(\frac{P_{n-1}(t)}{(1-t)^n})'+\beta_n\frac{P_{n-1}(t)}{(1-t)^n} \end{equation}
I don't understand how the author gets the equality above.
Let me write $$Q_n(t)=\frac{P_{n}(t)}{(1-t)^{n+1}}=\sum_{k\geq 0} f_n(k)t^k.$$ So then $$Q_{n-1}'(t)=\sum_{k\geq 0} kf_{n-1}(k)t^{k-1}$$ and $$tQ_{n-1}'(t)=\sum_{k\geq 0} kf_{n-1}(k)t^k$$ so $$\alpha_n tQ_{n-1}'(t)+\beta_nQ_{n-1}(t)=\sum_{k\geq 0} (\alpha_n k+\beta_n)f_{n-1}(k)t^k.$$ Assuming $f_n(k)=(\alpha_n k+\beta_n)f_{n-1}(k)$, the right-hand side is exactly $Q_n(t)$, so this says $$Q_n(t)=\alpha_n tQ_{n-1}'(t)+\beta_nQ_{n-1}(t)$$ which is the equality you are asking about.
(Note that for this we need $f_n(k)=(\alpha_n k+\beta_n)f_{n-1}(k)$, not $f_{n+1}(k)=(\alpha_n k+\beta_n)f_n(k)$ as you wrote. It appears your author made an indexing error here.)