closed form expression for the generating function $\sum_{n=0}^\infty \binom{m+n}{n} x^n$

Question

How to find the closed form expression for the generating function: $$ \sum_{n=0}^\infty \binom{m+n}{n} x^n\quad \mbox{where}\ m\ \mbox{is a}\ positive\ integer\ ?. $$

Answer 1

We are gonna show that for all $n\in\mathbb Z_{\geqslant 0}$,

$$f_n(x):=\sum_{n=0}^{+\infty}{m+n\choose n} x^n = \frac{1}{(1-x)^{m+1}}$$

For $m=0$, ${n \choose n}=1$ and therefore

$$\sum_{n=0}^{+\infty}{n\choose n}x^n =\sum_{n=0}^{+\infty}x^n =\frac{1}{1-x}=\frac{1}{(1-x)^{m+1}}$$

If the formula holds for some $m\geqslant 0$, then we're gonna show that it also holds for $m+1$. By Pascal's formula, $${m+1+n\choose n}={m+n \choose n}+{m+n \choose n-1}$$ Therefore, $$f_{m+1}(x)=\sum_{n=0}^{+\infty}{m+1+n\choose n}x^n =f_m(x)+ \sum_{n=0}^{+\infty}{m+n+1\choose n}x^{n+1}=f_m(x)+xf_{m+1}(x)$$ and you get the relation $$(1-x)f_{m+1}(x)=f_m(x)$$ and then $$f_{m+1}=\frac{f_m(x)}{1-x}=\frac{1}{(1-x)^{m+2}}$$ By induction the formula holds for all $m$.

Answer 2

$$ \color{#44f}{\Large\sum_{n = 0}^{\infty}{m + n \choose n}x^{n}} = \sum_{n = 0}^{\infty}{-m - 1 \choose n}\left(-1\right)^{n}x^{n} = \color{#44f}{\Large\left(1 - x\right)^{-m - 1}} $$