Dtjytyk

I am using an array environment to get aligned portions of a series of equations to center (instead of left-justify), as shown below:

usepackage{array,amsmath}

[

begin{array}{>{displaystyle}c @{{}={}} >{displaystyle}c @{{}+{}} >{displaystyle}c @{{}+{}} >{displaystyle}c}

sumlimits_{r=0}^{n+1} binom{n+1}{r} & binom{n+1}{0} & binom{n+1}{1} + ldots + binom{n+1}{n} & binom{n+1}{n+1} \

& 1 & sumlimits_{r=1}^n binom{n+1}{r} & 1 \

end{array}

]

The array environment (I believe) is necessary here to get each of the columns to center instead of justifying left.

Now my problem is that these two lines are part of a greater series of equations, where the others do not follow this pattern to be aligned. However, I need the equals signs to line up across all lines.

My current approach is follow the array with a normal align environment, having one equation line mirroring the longest line above but enclosed in phantom{} to get the align spacing right. But this leaves a single empty line with an equals in it.

...



begin{align*}

&= 2 + sum_{r=1}^nleft[binom{n}{r} + binom{n}{r-1}right] \

phantom{sumlimits_{r=0}^{n+1} binom{n+1}{r}} &= phantom{ binom{n+1}{0} + binom{n+1}{1} + ldots + binom{n+1}{n} + binom{n+1}{n+1}}

end{align*}

How can I get this result, but without the extraneous equals line at the end? Preferable a more elegant one, as this idea relies on several iffy factors such as none of the following equations exceeding the size of the one governing the special alignment.

eqparbox allows you to store the lengths of boxes via a <tag>. Boxes with the same <tag> are set with the maximum width across all content. Below I use this approach with a newly-defined eqmathbox[<tag>][<align>] (default for <align> is to centre the content) to add content to three different <tag>ged boxes:

documentclass{article}



usepackage{eqparbox,xparse,amsmath}



% https://tex.stackexchange.com/a/34412/5764

makeatletter

NewDocumentCommand{eqmathbox}{o O{c} m}{%

  IfValueTF{#1}

    {defeqmathbox@##1##2{eqmakebox[#1][#2]{$##1##2$}}}

    {defeqmathbox@##1##2{eqmakebox{$##1##2$}}}

  mathpaletteeqmathbox@{#3}

}

makeatother



begin{document}



begin{align*}

sum_{r = 0}^{n + 1} binom{n + 1}{r} 

    &= eqmathbox[LEFT]{binom{n + 1}{0}} + eqmathbox[CENTRE]{binom{n + 1}{1} + dots + binom{n + 1}{n}} + eqmathbox[RIGHT]{binom{n + 1}{n + 1}} \

    &= eqmathbox[LEFT]{1} + eqmathbox[CENTRE]{sum_{r = 1}^n binom{n + 1}{r}} + eqmathbox[RIGHT]{1} \

    &= 2 + sum_{r = 1}^n biggl[ binom{n}{r} + binom{n}{r - 1} biggr]

end{align*}



end{document}

Since eqparbox uses TeX's label-ref system, you need to compile twice for every change in the content of the maximum width.

try

documentclass{article}

usepackage{array,amsmath}

begin{document}

[

begin{array}{>{displaystyle}c @{{}={}} >{displaystyle}c @{{}+{}} >{displaystyle}c @{{}+{}} >{displaystyle}c}

sum_{r=0}^{n+1} binom{n+1}{r} 

    & binom{n+1}{0} & binom{n+1}{1} + ldots + binom{n+1}{n} & binom{n+1}{n+1} \

    & 1 & sumlimits_{r=1}^n binom{n+1}{r} & 1 \

    & multicolumn{3}{>{displaystyle}l}{

      2 + sum_{r=1}^nleft[binom{n}{r} + binom{n}{r-1}right]

                        }

end{array}

]

end{document}

Use the [t] option. Then you do not need to use multicolumn many times if you have many subsequent lines.

documentclass{article}

usepackage{array,amsmath}

begin{document}

begin{align*}

sumlimits_{r=0}^{n+1}  binom{n+1}{r}

&begin{array}[t]{@{}>{displaystyle}c @{{}={}}@{}>{displaystyle}c @{{}+{}} >{displaystyle}c @{{}+{}} >{displaystyle}c}

& binom{n+1}{0} & binom{n+1}{1} + ldots + binom{n+1}{n} & binom{n+1}{n+1} \

& 1 & sumlimits_{r=1}^n binom{n+1}{r} & 1 \

end{array}\

&=2 + sum_{r=1}^nleft[binom{n}{r} + binom{n}{r-1}right]

end{align*}

end{document}

I would take a different different approach to displaying the material and showing which parts are equal to what: I'd use three underbrace directives. I'd also use an align* environment.

documentclass{article}

usepackage{amsmath}

begin{document}



begin{align*}

sum_{r=0}^{n+1} binom{n+1}{r} 

&= {underbrace{binom{n+1}{0}}_{displaystyle 1}} 

 + {underbrace{binom{n+1}{1} + dots + binom{n+1}{n}}_{%

    displaystyle sum_{r=1}^n binom{n+1}{r}}} 

 + {underbrace{binom{n+1}{n+1}}_{displaystyle 1}} \

&= 2 + sum_{r=1}^n biggl[binom{n}{r} + binom{n}{r-1}biggr]

end{align*}



end{document}