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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?

Ben
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  • You may want to see this: https://math.stackexchange.com/questions/1784613/another-hockey-stick-identity. (Replace their $r$ with your $n$ and their $n$ with your $m+1$.) – Minus One-Twelfth Mar 22 '19 at 03:17
  • This is easy to show by induction ... try it ? – Donald Splutterwit Mar 22 '19 at 03:22
  • Alternative answer ... Yes someone can explain it ! – Donald Splutterwit Mar 22 '19 at 03:23
  • @DonaldSplutterwit I know that it's done by induction but still sure how. So yes, I've tried it. – Ben Mar 22 '19 at 04:05
  • @MinusOne-Twelfth It's almost impossible to keep the terms straight but I substituted my stuff over to the best reply on that post but it still isn't very clear. Do you know the induction method? – Ben Mar 22 '19 at 04:06
  • Try inducting on $n$. The main part is (the inductive step) assuming that $\sum\limits_{i=0}^{n}\binom{m+i}{m} = \binom{m+n+1}{m+1}$ and showing that $\sum\limits_{i=0}^{n+1}\binom{m+i}{m} = \binom{m+n+2}{m+1}$ (I used $\binom{a}{b} = \binom{a}{a-b}$ in some places so the lower number in the binomial coefficients may not look the same as in your original post). You may also want to recall Pascal's rule to help you. – Minus One-Twelfth Mar 22 '19 at 04:15
  • If you think of it in terms of "Pascal"'s triangle, it's obvious. It's the sum of a 'diagonal' starting with 1. Use the fact that the sum of two adjacent terms gives the term on the next row. – Chrystomath Mar 22 '19 at 05:08

2 Answers2

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

Felix Marin
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We have

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

Marko Riedel
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