# Pascal's triangle with general (2n)-th row

I am interested in creating Pascal's triangle as in this answer for N=6, but add the general (2n)-th row showing the first binomial coefficient, then dots, then the 3 middle binomial coefficients, then dots, then the last one.

Is this possible? I am very new to tikz and therefore happy to receive any kind of tip to solve this. Thank you!

Welcome! This is to give you a start.

\documentclass[tikz,border=3mm]{standalone}
\usepackage{amsmath}
\begin{document}
\begin{tikzpicture}[x=0.75cm,y=0.5cm,
pascal node/.style={font=\footnotesize},
row node/.style={font=\footnotesize, anchor=west, shift=(180:1)},
Dotted/.style={% https://tex.stackexchange.com/a/52856/194703
dash pattern=on 0.1\pgflinewidth off #1\pgflinewidth,line cap=round,
shorten >=2pt,shorten <=2pt,line width=1pt},
Dotted/.default=4,thick]
\def\NPascal{6}
\def\offset{2}
\draw
foreach \n in {0,...,\NPascal} {
(-\NPascal/2-1-\offset, -\n) node (rn-\n) [row node/.try]{Row \n:}
\foreach \k in {0,...,\n}{
(-\n/2+\k,-\n) node (pn-\n-\k) [pascal node/.try] {%
$\binom{\n}{\k}$
}}}
(-\NPascal/2-1-\offset, -\NPascal-2) node (rn-N) [row node/.try]{Row $2n$:}
(-\NPascal/2-1, -\NPascal-2) node (pn-N-0) [pascal node/.try] {$\binom{2n}{0}$}
(\NPascal/2+1, -\NPascal-2) node (pn-N-N) [pascal node/.try] {$\binom{2n}{2n}$}
(rn-\NPascal) edge[Dotted] (rn-N.north-|rn-\NPascal)
(pn-\NPascal-0) edge[Dotted] (pn-N-0)
(pn-\NPascal-\NPascal) edge[Dotted] (pn-N-N)
(pn-N-N) edge[Dotted] (pn-N-0);
\end{tikzpicture}
\end{document}