Exempt portion of equation line from aligning?












3















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}
]


eqn1



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*}


eqn2



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.










share|improve this question







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    3















    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}
    ]


    eqn1



    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*}


    eqn2



    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.










    share|improve this question







    New contributor




    PGmath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.























      3












      3








      3








      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}
      ]


      eqn1



      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*}


      eqn2



      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.










      share|improve this question







      New contributor




      PGmath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.












      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}
      ]


      eqn1



      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*}


      eqn2



      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.







      math-mode horizontal-alignment align arrays






      share|improve this question







      New contributor




      PGmath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|improve this question







      New contributor




      PGmath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|improve this question




      share|improve this question






      New contributor




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      Check out our Code of Conduct.









      asked 38 mins ago









      PGmathPGmath

      1162




      1162




      New contributor




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      New contributor





      PGmath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






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      Check out our Code of Conduct.






















          3 Answers
          3






          active

          oldest

          votes


















          1














          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}


          enter image description here






          share|improve this answer































            1














            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}


            enter image description here






            share|improve this answer


























            • I like this approach better but I see it misses the equals on the 2nd line.

              – PGmath
              18 mins ago













            • @PGmath Very good catch! My bad. I updated.

              – marmot
              15 mins ago











            • Thanks. Can you explain a little what [t] does? I've never done much involved stuff with arrays before.

              – PGmath
              just now



















            1














            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:



            enter image description here



            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.






            share|improve this answer

























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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              1














              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}


              enter image description here






              share|improve this answer




























                1














                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}


                enter image description here






                share|improve this answer


























                  1












                  1








                  1







                  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}


                  enter image description here






                  share|improve this answer













                  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}


                  enter image description here







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 28 mins ago









                  ZarkoZarko

                  126k868165




                  126k868165























                      1














                      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}


                      enter image description here






                      share|improve this answer


























                      • I like this approach better but I see it misses the equals on the 2nd line.

                        – PGmath
                        18 mins ago













                      • @PGmath Very good catch! My bad. I updated.

                        – marmot
                        15 mins ago











                      • Thanks. Can you explain a little what [t] does? I've never done much involved stuff with arrays before.

                        – PGmath
                        just now
















                      1














                      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}


                      enter image description here






                      share|improve this answer


























                      • I like this approach better but I see it misses the equals on the 2nd line.

                        – PGmath
                        18 mins ago













                      • @PGmath Very good catch! My bad. I updated.

                        – marmot
                        15 mins ago











                      • Thanks. Can you explain a little what [t] does? I've never done much involved stuff with arrays before.

                        – PGmath
                        just now














                      1












                      1








                      1







                      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}


                      enter image description here






                      share|improve this answer















                      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}


                      enter image description here







                      share|improve this answer














                      share|improve this answer



                      share|improve this answer








                      edited 15 mins ago

























                      answered 21 mins ago









                      marmotmarmot

                      107k5129243




                      107k5129243













                      • I like this approach better but I see it misses the equals on the 2nd line.

                        – PGmath
                        18 mins ago













                      • @PGmath Very good catch! My bad. I updated.

                        – marmot
                        15 mins ago











                      • Thanks. Can you explain a little what [t] does? I've never done much involved stuff with arrays before.

                        – PGmath
                        just now



















                      • I like this approach better but I see it misses the equals on the 2nd line.

                        – PGmath
                        18 mins ago













                      • @PGmath Very good catch! My bad. I updated.

                        – marmot
                        15 mins ago











                      • Thanks. Can you explain a little what [t] does? I've never done much involved stuff with arrays before.

                        – PGmath
                        just now

















                      I like this approach better but I see it misses the equals on the 2nd line.

                      – PGmath
                      18 mins ago







                      I like this approach better but I see it misses the equals on the 2nd line.

                      – PGmath
                      18 mins ago















                      @PGmath Very good catch! My bad. I updated.

                      – marmot
                      15 mins ago





                      @PGmath Very good catch! My bad. I updated.

                      – marmot
                      15 mins ago













                      Thanks. Can you explain a little what [t] does? I've never done much involved stuff with arrays before.

                      – PGmath
                      just now





                      Thanks. Can you explain a little what [t] does? I've never done much involved stuff with arrays before.

                      – PGmath
                      just now











                      1














                      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:



                      enter image description here



                      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.






                      share|improve this answer






























                        1














                        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:



                        enter image description here



                        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.






                        share|improve this answer




























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                          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:



                          enter image description here



                          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.






                          share|improve this answer















                          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:



                          enter image description here



                          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.







                          share|improve this answer














                          share|improve this answer



                          share|improve this answer








                          edited 3 mins ago

























                          answered 15 mins ago









                          WernerWerner

                          446k699871692




                          446k699871692






















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