Prove $\sum_{i=0}^n\binom{m+i}{i}=\binom{m+n+1}{n}$ (another Hockey-Stick Identity?)

Question

Let $n$ be a nonnegative integer, and $m$ a positive integer. Could someone explain to me why the identity $$ \sum_{i=0}^n\binom{m+i}{i}=\binom{m+n+1}{n} $$ holds?

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\sum_{i = 0}^{n}{m + i \choose i}} = \sum_{i = 0}^{n}\pars{-1}^{i}{-m -i + i - 1 \choose i} \\[5mm] = &\ \sum_{i = 0}^{n}\pars{-1}^{i}{-m - 1 \choose i} = \sum_{i = 0}^{n}\pars{-1}^{i}\bracks{z^{i}}\pars{1 + z}^{-m - 1} \\[5mm] = &\ \bracks{z^{0}}\pars{1 + z}^{-m - 1} \sum_{i = 0}^{n}\pars{-\,{1 \over z}}^{i} = \bracks{z^{0}}\pars{1 + z}^{-m - 1}\, {\pars{-1/z}^{n + 1} - 1 \over \pars{-1/z} - 1} \\[5mm] = &\ \bracks{z^{0}}\pars{1 + z}^{-m - 1}\, {\pars{-1}^{n + 1} - z^{n + 1} \over -1 - z} \,{z \over z^{n + 1}} \\[5mm] = &\ \bracks{z^{n}}\pars{1 + z}^{-m - 2}\, \bracks{z^{n + 1} - \pars{-1}^{n + 1}} = \pars{-1}^{n}\bracks{z^{n}}\pars{1 + z}^{-m - 2} \\[5mm] = &\ \pars{-1}^{n}{-m - 2 \choose n} = \pars{-1}^{n}\bracks{{m + 2 + n - 1\choose n}\pars{-1}^{n}} \\[5mm] = &\ \bbx{m + n + 1 \choose n} \\ & \end{align}

Answer 2

We have

$$\sum_{q=0}^n {m+q\choose q} = \sum_{q\ge 0} {m+q\choose q} [[0\le q\le n]] \\ = \sum_{q\ge 0} {m+q\choose q} [z^n] \frac{z^q}{1-z} = [z^n] \frac{1}{1-z} \sum_{q\ge 0} {m+q\choose q} z^q \\= [z^n] \frac{1}{1-z} \frac{1}{(1-z)^{m+1}} = [z^n] \frac{1}{(1-z)^{m+2}} = {n+m+1\choose n}.$$