1
$\begingroup$

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?

$\endgroup$

closed as off-topic by Hanul Jeon, Leucippus, José Carlos Santos, Eevee Trainer, Parcly Taxel Mar 26 at 3:27

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Hanul Jeon, Leucippus, José Carlos Santos, Eevee Trainer, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ You may want to see this: math.stackexchange.com/questions/1784613/…. (Replace their $r$ with your $n$ and their $n$ with your $m+1$.) $\endgroup$ – Minus One-Twelfth Mar 22 at 3:17
  • $\begingroup$ This is easy to show by induction ... try it ? $\endgroup$ – Donald Splutterwit Mar 22 at 3:22
  • $\begingroup$ Alternative answer ... Yes someone can explain it ! $\endgroup$ – Donald Splutterwit Mar 22 at 3:23
  • $\begingroup$ @DonaldSplutterwit I know that it's done by induction but still sure how. So yes, I've tried it. $\endgroup$ – Ben Mar 22 at 4:05
  • $\begingroup$ @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? $\endgroup$ – Ben Mar 22 at 4:06
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ The color #ffd is so ease to work with :) $\endgroup$ – Le Anh Dung Mar 22 at 5:52
  • 2
    $\begingroup$ @LeAnhDung It clearly shows the very beginning while the box shows the very end. $\endgroup$ – Felix Marin Mar 22 at 5:54
2
$\begingroup$

We have

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

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.