TITLE: how to prove a combinatorial identity
QUESTION [5 upvotes]: I've encountered with following identity:
$$\sum\limits_{n=0}^\infty\binom{a+bn}{n} \left(\frac{z}{(1+z)^b}\right)^n=\frac{(1+z)^{1+a}}{1+(1-b)z}$$
Is it correct? how to prove it?

REPLY [2 votes]: By using Lagrange's expansion
\begin{align}
\frac{f(z)}{1 - w \phi(z)} = \sum_{n=0}^{\infty} \frac{w^{n}}{n!} \, \left. D_{z}^{n} \left\{ f(z) \, [ \phi(z) ]^{n} \right\} \right|_{z = z_{0}}
\end{align}
where $z = z_{0} + w \phi(z)$ then one easily developes 
\begin{align}
\sum_{n=0}^{\infty} \binom{a+bn}{n} \, \left(\frac{z}{(1+z)^b}\right)^{n} = \frac{(1+z)^{1+a}}{1+(1-b)z}
\end{align}