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0

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

share

asked 1 min ago

PGmathPGmath

1011

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.

add a comment | 

0

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

share

asked 1 min ago

PGmathPGmath

1011

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.

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

share

asked 1 min ago

PGmathPGmath

1011

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.

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}

]

...



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*}

share

asked 1 min ago

PGmathPGmath

1011

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

asked 1 min ago

PGmathPGmath

1011

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.

asked 1 min ago

PGmathPGmath

1011

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.

asked 1 min ago

PGmathPGmath

1011