Exempt portion of equation line from aligning?
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.
math-mode horizontal-alignment align arrays
add a comment |
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.
math-mode horizontal-alignment align arrays
add a comment |
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.
math-mode horizontal-alignment align arrays
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.
math-mode horizontal-alignment align arrays
math-mode horizontal-alignment align arrays
asked Mar 9 at 4:00
PGmathPGmath
20315
20315
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
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 like this approach better but I see it misses the equals on the 2nd line.
– PGmath
Mar 9 at 4:20
@PGmath Very good catch! My bad. I updated.
– marmot
Mar 9 at 4:24
Thanks. Can you explain a little what[t]
does? I've never done much involved stuff with arrays before.
– PGmath
Mar 9 at 4:39
@PGmath It aligns the array at the top.
– marmot
Mar 9 at 4:39
add a comment |
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 c
entre 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.
add a comment |
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}
add a comment |
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}
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 like this approach better but I see it misses the equals on the 2nd line.
– PGmath
Mar 9 at 4:20
@PGmath Very good catch! My bad. I updated.
– marmot
Mar 9 at 4:24
Thanks. Can you explain a little what[t]
does? I've never done much involved stuff with arrays before.
– PGmath
Mar 9 at 4:39
@PGmath It aligns the array at the top.
– marmot
Mar 9 at 4:39
add a comment |
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 like this approach better but I see it misses the equals on the 2nd line.
– PGmath
Mar 9 at 4:20
@PGmath Very good catch! My bad. I updated.
– marmot
Mar 9 at 4:24
Thanks. Can you explain a little what[t]
does? I've never done much involved stuff with arrays before.
– PGmath
Mar 9 at 4:39
@PGmath It aligns the array at the top.
– marmot
Mar 9 at 4:39
add a comment |
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}
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}
edited Mar 9 at 4:23
answered Mar 9 at 4:17
marmotmarmot
117k5150283
117k5150283
I like this approach better but I see it misses the equals on the 2nd line.
– PGmath
Mar 9 at 4:20
@PGmath Very good catch! My bad. I updated.
– marmot
Mar 9 at 4:24
Thanks. Can you explain a little what[t]
does? I've never done much involved stuff with arrays before.
– PGmath
Mar 9 at 4:39
@PGmath It aligns the array at the top.
– marmot
Mar 9 at 4:39
add a comment |
I like this approach better but I see it misses the equals on the 2nd line.
– PGmath
Mar 9 at 4:20
@PGmath Very good catch! My bad. I updated.
– marmot
Mar 9 at 4:24
Thanks. Can you explain a little what[t]
does? I've never done much involved stuff with arrays before.
– PGmath
Mar 9 at 4:39
@PGmath It aligns the array at the top.
– marmot
Mar 9 at 4:39
I like this approach better but I see it misses the equals on the 2nd line.
– PGmath
Mar 9 at 4:20
I like this approach better but I see it misses the equals on the 2nd line.
– PGmath
Mar 9 at 4:20
@PGmath Very good catch! My bad. I updated.
– marmot
Mar 9 at 4:24
@PGmath Very good catch! My bad. I updated.
– marmot
Mar 9 at 4:24
Thanks. Can you explain a little what
[t]
does? I've never done much involved stuff with arrays before.– PGmath
Mar 9 at 4:39
Thanks. Can you explain a little what
[t]
does? I've never done much involved stuff with arrays before.– PGmath
Mar 9 at 4:39
@PGmath It aligns the array at the top.
– marmot
Mar 9 at 4:39
@PGmath It aligns the array at the top.
– marmot
Mar 9 at 4:39
add a comment |
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 c
entre 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.
add a comment |
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 c
entre 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.
add a comment |
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 c
entre 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.
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 c
entre 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.
edited Mar 9 at 4:35
answered Mar 9 at 4:24
WernerWerner
450k729991709
450k729991709
add a comment |
add a comment |
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}
add a comment |
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}
add a comment |
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}
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}
answered Mar 9 at 4:11
ZarkoZarko
129k868169
129k868169
add a comment |
add a comment |
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}
add a comment |
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}
add a comment |
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}
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}
answered Mar 9 at 9:49
MicoMico
286k32391779
286k32391779
add a comment |
add a comment |
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