how to prove a combinatorial identity

Question

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?

Answer 1

Let $~S(a,b)=\displaystyle\sum_{n=0}^\infty{a+bn\choose n}~x^n.\quad$ Then $~S(a,\color{red}0)=(1+x)^a,\quad S(a,\color{red}1)=\dfrac1{(1-x)^{a+1}}~,$

$S(0,\color{red}2)=\dfrac1{\sqrt{1-4x}}~,~$ and $~S(a,\color{red}3)$ yields a combination of algebraic and trigonometric

functions. For $b\ge\color{red}4$, there do not seem to be any meaningful closed forms, except for

rewriting the sum in terms of hypergeometric functions.

Answer 2

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}