5
  1. A series of equally-spaced dots are required to complete the set up.

  2. A slight adjustment of the vertical alignment is required.

enter image description here

The above set up must appear as something like this:-

enter image description here

\documentclass[11pt,a4paper]{article}
\usepackage{amsmath}
\usepackage{array}
\usepackage{booktabs}

\begin{document}
using the detached coefficients, we have the scheme\\\\
\noindent
$\begin{array}{@{} >{{}}r @{} >{{}}r *{20}{ @{}>{{}}r<{{}}@{} } }
& 1 & + & _nC_{1} & + & _nC_{2} & + & _nC_{3} & + & \dotsb  & + & _nC_{n-1} & + & _nC_{n} \\
\times \\
& 1 & + &  1      & \\
\midrule
& 1 &   &         & + & _nC_{1} &   &         & + & _nC_{2} &   &   &   &   &   &   &         & + & _nC_{n} \\
&   &   &         & + &  1      &   &         & + & _nC_{1} &   &   &   &   &   &   &      & + & _nC_{n-1} & + & _nC_{n} \\
\midrule
\end{array}$
\end{document}

1 Answer 1

5

You can use \hdotsfor from amsmath (the argument indicates the number of columns to be spanned):

\documentclass[11pt,a4paper]{article}
\usepackage{amsmath}
\usepackage{array}
\usepackage{booktabs}

\begin{document}
using the detached coefficients, we have the scheme\\\\
\noindent
$\begin{array}{@{} >{{}}r @{} >{{}}r *{20}{ @{}>{{}}r<{{}}@{} } }
& 1 & + & _nC_{1} & + & _nC_{2} & + & _nC_{3} & + & \dotsb  & + & _nC_{n-1} & + & _nC_{n} \\
\times \\
& 1 & + &  1      & \\
\midrule
& 1 &   &         & + & _nC_{1} & \hdotsfor{2}          & + & _nC_{2} &   \hdotsfor{3}           + & _nC_{n} \\
&   &   &         & + &  1  & \hdotsfor{2}         & + & _nC_{1} &  \hdotsfor{3}       + & _nC_{n-1} & + & _nC_{n} \\
\midrule
\end{array}$
\end{document}

enter image description here

With l (left) aligned columns strating from the second one:

\documentclass[11pt,a4paper]{article}
\usepackage{amsmath}
\usepackage{array}
\usepackage{booktabs}

\begin{document}
using the detached coefficients, we have the scheme\\\\
\noindent
$\begin{array}{@{} >{{}}r @{} >{{}}r *{20}{ @{}>{{}}l<{{}}@{} } }
& 1 & + & _nC_{1} & + & _nC_{2} & + & _nC_{3} & + & \dotsb  & + & _nC_{n-1} & + & _nC_{n} \\
\times \\
& 1 & + &  1      & \\
\midrule
& 1 &   &         & + & _nC_{1} & \hdotsfor{2}          & + & _nC_{2} &   \hdotsfor{3}           + & _nC_{n} \\
&   &   &         & + &  1  & \hdotsfor{2}         & + & _nC_{1} &  \hdotsfor{3}       + & _nC_{n-1} & + & _nC_{n} \\
\midrule
\end{array}$
\end{document}

enter image description here

18
  • Thank you very much for the improvement! I really appreciate it. Would it be difficult for you to align the set up as in the second figure? Example:- Of the second row, + and 1 are quite distant apart. Mar 4, 2015 at 16:48
  • 1
    @NisalKevinKotinkaduwa just change all the r but the first one to l. I am talking about your \begin{array}{...} definition
    – LaRiFaRi
    Mar 4, 2015 at 16:50
  • @LaRiFaRi Are you suggesting me to replace the first r by 1? Mar 4, 2015 at 16:56
  • 1
    @NisalKevinKotinkaduwa No, not the first one. The other one(s), where left alignment is desired. Mar 4, 2015 at 16:57
  • 1
    @NisalKevinKotinkaduwa You're welcome! I'm glad I could help. Mar 4, 2015 at 17:44

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