Show $\sum_{i=0}^\infty {k+i \choose k}a^i=\frac1{(1-a)^{k+1}}.$ I need to show that for every $k\in\mathbb{N}, |a|<1,$ $$\sum_{i=0}^\infty {k+i \choose k}a^i=\frac1{(1-a)^{k+1}}.$$ 
It's technically a power series in $a$ but no approach in that direction proved fruitful.
My only ideas are that $\sum_{i=0}^\infty i^{k+1}a^i=\frac{p(a)}{(1-a)^{k+1}}$ for some polynomial $p$, and that $${k+i \choose k}=\frac{(i+1)\cdots(i+k)}{1\cdots k}.$$
 A: Try induction on $k$.
For $k=0$, we have $\sum_{i=0}^\infty a^i = \frac{1}{1-a}$
For $k>0$ we can write $\binom{k+i}{k} = \binom{k+i-1}{k-1} + \binom{k+i-1}{k}$, so 
$$
\begin{align}
\sum_{i=0}^\infty \binom{k+i}{k}a^i &= \sum_{i=0}^\infty \binom{k+i-1}{k-1}a^i + \sum_{i=0}^\infty \binom{k+i-1}{k}a^i \\
&= \frac{1}{(1-a)^k} + \sum_{i=1}^\infty \binom{k+i-1}{k}a^i \\
&= \frac{1}{(1-a)^k} + \sum_{i=0}^\infty \binom{k+i}{k}a^{i+1} \\
(1-a)\sum_{i=0}^\infty \binom{k+i}{k}a^i &= \frac{1}{(1-a)^k} \\
\sum_{i=0}^\infty \binom{k+i}{k}a^i &= \frac{1}{(1-a)^{k+1}}
\end{align}
$$
A: Let $f(a)=(1-a)^{-k-1}$. Then


*

*$f'(a)=(k+1)(1-a)^{-k-2}$;

*$f''(a)=(k+1)(k+2)(1-a)^{-k-3}$


and so on. In fact, if $n\in\mathbb N$,$$f^{(n)}(a)=(k+1)(k+2)\cdots(k+n)(1-a)^{-k-n-1}$$an therefore$$\frac{f^{(n)}(0)}{n!}=\binom{k+n}k.$$So$$\bigl(\forall a\in(-1,1)\bigr):\frac1{(1-a)^{k+1}}=\sum_{n=0}^\infty\binom{k+n}ka^n,$$because the radius of convergence of this power series is $1$ and because $\frac1{(1-a)^{k+1}}$ is an analytic function.
A: That's a classical entry in the tables for z-Transform.
Note that , by applying "Symmetry" and then "Upper-negation" to the binomial coefficient we get
$$
\left( \matrix{
  k + i \cr 
  k \cr}  \right)\quad \left| {\;0 \le k,i \in Z} \right.\quad  = \left( \matrix{
  k + i \cr 
  i \cr}  \right) = \left( { - 1} \right)^{\,i} \left( \matrix{
   - \left( {k + 1} \right) \cr 
  i \cr}  \right)
$$
So
$$
\sum\limits_{0\, \le \,i} {\left( \matrix{
  k + i \cr 
  k \cr}  \right)a^{\,i} }  = \sum\limits_{0\, \le \,i} {\left( \matrix{
   - \left( {k + 1} \right) \cr 
  i \cr}  \right)\left( { - a} \right)^{\,i} }  = \left( {1 - a} \right)^{\, - \,\left( {k + 1} \right)} 
$$
is reconducted under the "generalized binomial expansion", and converges for $|a|<1$.
A: Hint: the geometric series says that
$$\frac{1}{1-a} = \sum_{i=0}^{\infty} a^i. $$
Try taking successive derivatives of both sides. (Ask yourself why this works and why I picked this function in the first place.) 
A: Copied from this answer to What does $\binom{-n}{k}$ mean?:

It is the binomial coefficient for a negative exponent:
  $$
\begin{align}
(1+x)^{-n}
&=\sum_{k=0}^\infty\binom{-n}{k}x^k\\
&=\sum_{k=0}^\infty(-1)^k\binom{k+n-1}{k}x^k
\end{align}
$$
  Note that this follows from the following formulation of the standard binomial coefficient:
  $$
\begin{align}
\binom{-n}{k}
&=\frac{\overbrace{-n(-n-1)(-n-2)\dots(-n-k+1)}^{k\text{ factors}}}{k!}\\
&=(-1)^k\frac{(n+k-1)(n+k-2)(n+k-3)\dots n}{k!}\\
&=(-1)^k\binom{n+k-1}{k}
\end{align}
$$

A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\sum_{i = 0}^{\infty}{k + i \choose k}a^{i} & =
\sum_{i = 0}^{\infty}{k + i \choose i}a^{i} =
\sum_{i = 0}^{\infty}\bracks{{-k - 1 \choose i}\pars{-1}^{i}}a^{i} =
\sum_{i = 0}^{\infty}{-k - 1 \choose i}\pars{-a}^{i}
\\[5mm] & =
\bracks{1 + \pars{-a}}^{-k - 1} = \bbx{1 \over \pars{1 - a}^{k + 1}}
\end{align}
