1

I'm having a difficult time understanding how to give a combinatorics proof of the identity $$\sum_{k=0}^{n} \binom{x+k}{k}=\binom{x+n+1}{n}$$

RobPratt
  • 45,619
VN4
  • 51
  • edited title and body, thanks! – VN4 Aug 30 '20 at 00:29
  • Possibly not relevant now, but the answer to your original question was $$\sum_{k=0}^{n} \binom{x+k}{n}=\binom{x+n+1}{n+1}-\binom{x}{n+1}.$$ This can be proved by noting that both the LHS and the RHS count the number of subsets of ${1,2,\cdots,x+n}$ of size $n$, together with an integer $k\in {0,1,2,\cdots n}$ such that $x+k$ is at least the largest element of the subset. – tkf Aug 30 '20 at 00:43

2 Answers2

1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{\sum_{k = 0}^{n}{x + k \choose k} = {x + n + 1 \choose n}}:\ {\Large ?}}$.

\begin{align} \bbox[5px,#ffd]{\sum_{k = 0}^{n}{x + k \choose k}} & = \sum_{k = 0}^{n}\bracks{z^{k}}\pars{1 + z}^{x + k} = \bracks{z^{0}}\pars{1 + z}^{x}\sum_{k = 0}^{n}\pars{1 + z \over z}^{k} \\[5mm] & = \bracks{z^{0}}\pars{1 + z}^{x}\, {\pars{1 + z}^{n + 1}/z^{n + 1} - 1 \over \pars{1 + z}/z - 1} \\[5mm] = &\ \bracks{z^{0}}\pars{1 + z}^{x}\,{1 \over z^{n}} \bracks{\pars{1 + z}^{n + 1} - z^{n + 1}} \\[5mm] = &\ \bracks{z^{n}}\pars{1 + z}^{x + n + 1} = \bbx{\large{x + n + 1 \choose n}} \\ & \end{align}
Felix Marin
  • 89,464
0

This can be proved by noting that the sum counts the number of subsets of $\{1,2,\cdots,x+n\}$ of size $x$, together with an integer $k\in \{0,1,2,\cdots n\}$ such that $x+k$ is at least the largest element of the subset. Seperating into the two cases where $k$ equals the largest element, and where it is strictly greater than the largest element, we get: \begin{eqnarray*}\sum_{k=0}^{n} \binom{x+k}{k}&=&\sum_{k=0}^{n} \binom{x+k}{x}\\ &=&\binom{x+n}{x}+\binom{x+n}{x+1}\\&=&\binom{x+n+1}{x+1}\\&=&\binom{x+n+1}{n}.\end{eqnarray*}

tkf
  • 11,563