Why can I not vertically align the text in the 3rd table column?

I'm trying to make a table which lists some math functions. I've been searching the net on how to center the equations in the cells both vertically and horizontally. And I think I've found an answer to it, except that no matter what I do the equations in the 3rd column won't align, and just stays on the top. I use the array package. What am I doing wrong?

    \begin{table}[H]
\centering
\newcolumntype{A}{ >{\centering\arraybackslash} m{1cm} }
\newcolumntype{B}{ >{\centering\arraybackslash} m{4cm} }
\newcolumntype{C}{ >{\centering\arraybackslash} m{2cm} }
\begin{tabular}{|A|B|C|}
\hline

$n$ & $a_{n}$ & $\frac{k_{0}^{n}}{n!}a_{0}$ \2ex] \hline 1 & k_{0}a_{0} & \frac{k_{0}^{1}}{1}a_{0} \\[2ex] \hline 2 & \frac{k_{0}}{2}a_{1}=\frac{k_{0}}{2}k_{0}a_{0} & \frac{k_{0}^{2}}{2 \cdot 1}a_{0} \\[2ex] \hline 3 & \frac{k_{0}}{3}a_{2}=\frac{k_{0}}{3}\frac{k_{0}}{2}k_{0}a_{0} & \frac{k_{0}^{3}}{3 \cdot 2 \cdot 1}a_{0} \\[2ex] \hline 4 & \frac{k_{0}}{4}a_{3}=\frac{k_{0}}{4}\frac{k_{0}}{3} \frac{k_{0}}{2}k_{0}a_{0} & \frac{k_{0}^{4}}{4 \cdot 3 \cdot 2 \cdot 1}a_{0} \\[2ex] \hline \end{tabular} \end{table}  - add comment 4 Answers For comparison, I think that ConTeXt provides a cleaner solution in this case. • Use a setup to specify the width of the three columns (width=...), specify that all cells should be horizontally and vertically middle aligned (align={middle,lohi}), and specify that each cell should have a 1ex top and bottom offset (toffset=1ex, boffset=1ex): \startsetups table:align \setupTABLE[align={middle,lohi}, toffset=1ex, boffset=1ex] \setupTABLE[column][1][width=1cm] \setupTABLE[column][2][width=4cm] \setupTABLE[column][3][width=2cm] \stopsetups  • Then use that setup for the tabular data \startTABLE[setups={table:align}] \NC n \NC a_{n} \NC \frac{k_{0}^{n}}{n!}a_{0} \NC \NR \NC 1 \NC k_{0}a_{0} \NC \frac{k_{0}^{1}}{1}a_{0} \NC \NR \NC 2 \NC \frac{k_{0}}{2}a_{1}=\frac{k_{0}}{2}k_{0}a_{0} \NC \frac{k_{0}^{2}}{2 \cdot 1}a_{0} \NC \NR \NC 3 \NC \frac{k_{0}}{3}a_{2}=\frac{k_{0}}{3}\frac{k_{0}}{2}k_{0}a_{0} \NC \frac{k_{0}^{3}}{3 \cdot 2 \cdot 1}a_{0} \NC \NR \NC 4 \NC \frac{k_{0}}{4}a_{3}=\frac{k_{0}}{4}\frac{k_{0}}{3} \frac{k_{0}}{2}k_{0}a_{0} \NC \frac{k_{0}^{4}}{4 \cdot 3 \cdot 2 \cdot 1}a_{0} \NC \NR \stopTABLE  - Thanks, I really appreciate it – Møller Nov 27 '12 at 18:20 add comment I would use an array environment nested inside an math environment of some sort: {\renewcommand{\arraystretch}{3} \[\begin{array}{|l|>{\displaystyle{}}c|>{\displaystyle{}}c|}\hline n & a_{n} & \frac{k_{0}^{n}}{n!}a_{0} \\\hline 1 & k_{0}a_{0} & \frac{k_{0}^{1}}{1}a_{0} \\\hline 2 & \frac{k_{0}}{2}a_{1} =\frac{k_{0}}{2}k_{0}a_{0} & \frac{k_{0}^{2}}{2 \cdot 1}a_{0} \\\hline 3 & \frac{k_{0}}{3}a_{2}=\frac{k_{0}}{3}\frac{k_{0}}{2}k_{0}a_{0} & \frac{k_{0}^{3}}{3 \cdot 2 \cdot 1}a_{0} \\\hline 4 & \frac{k_{0}}{4}a_{3}=\frac{k_{0}}{4}\frac{k_{0}}{3}\frac{k_{0}}{2}k_{0}a_{0} & \frac{k_{0}^{4}}{4 \cdot 3 \cdot 2 \cdot 1}a_{0} \\\hline \end{array}


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Ah great! Nice and simple, just the way I like it. Thank you very much –  Møller Nov 27 '12 at 18:21

I suspect it isn't your fault...

You can avoid the problem by making sure the m is not on the last column:

Also please always make your questions contain complete documents showing all packages used:

\documentclass{article}
\usepackage{array}
\setlength\extrarowheight{7pt}
\begin{document}
\centering
\newcolumntype{A}{ >{\centering\arraybackslash} m{1cm} }
\newcolumntype{B}{ >{\centering\arraybackslash} m{4cm} }
\newcolumntype{C}{ >{\centering\arraybackslash} m{2cm} }
\begin{tabular}{|A|B|C|@{}c@{}}
\hline

$n$ & $a_{n}$ & $\frac{k_{0}^{n}}{n!}a_{0}$ &\\[2ex]

\hline

1 & $k_{0}a_{0}$ & $\frac{k_{0}^{1}}{1}a_{0}$ &\\[2ex]

\hline

2 & $\frac{k_{0}}{2}a_{1}=\frac{k_{0}}{2}k_{0}a_{0}$
& $\frac{k_{0}^{2}}{2 \cdot 1}a_{0}$ &\\[2ex]

\hline

3 & $\frac{k_{0}}{3}a_{2}=\frac{k_{0}}{3}\frac{k_{0}}{2}k_{0}a_{0}$
& $\frac{k_{0}^{3}}{3 \cdot 2 \cdot 1}a_{0}$ &\\[2ex]

\hline

4 & $\frac{k_{0}}{4}a_{3}=\frac{k_{0}}{4}\frac{k_{0}}{3} \frac{k_{0}}{2}k_{0}a_{0}$
& $\frac{k_{0}^{4}}{4 \cdot 3 \cdot 2 \cdot 1}a_{0}$ &\\[2ex]

\hline
\end{tabular}

\end{document}

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Thank you sir :) I will do my best to include everything NeXT time I –  Møller Nov 27 '12 at 18:23
run in to problems –  Møller Nov 27 '12 at 18:23
feel free to ask another question with an example showing the problem –  David Carlisle Nov 27 '12 at 20:45

Given that it would appear that you're trying to make sure that math material is centered vertically, I think it may be a good idea not to insert something like [2ex] of extra vertical whitespace and, instead, insert "math struts" on each of the lines. The following MWE defines a macro named \RTS -- short for "really tall strut" -- that should be inserted somewhere in each row whose height you want to see enlarged.

\documentclass{article}
\usepackage{array,float}
\newcolumntype{A}{ >{\centering\arraybackslash} p{1cm} }
\newcolumntype{B}{ >{\centering\arraybackslash} p{4cm} }
\newcolumntype{C}{ >{\centering\arraybackslash} p{2cm} }
% RTS is short for "really tall strut":
\newcommand\RTS{$\vphantom{\int\limits_0^1}$}
\begin{document}
\begin{table}[H]
\centering
\begin{tabular}{|A|B|C|}
\hline
$n$ & $a_{n}$ & $\frac{k_{0}^{n}}{n!}a_{0}$ \RTS \\
\hline
1
& $k_{0}a_{0}$
& $\frac{k_{0}^{1}}{1}a_{0}$ \RTS \\
\hline
2
& $\frac{k_{0}}{2}a_{1}=\frac{k_{0}}{2}k_{0}a_{0}$
& $\frac{k_{0}^{2}}{2 \cdot 1}a_{0}$ \RTS \\
\hline
3 &
$\frac{k_{0}}{3}a_{2}=\frac{k_{0}}{3}\frac{k_{0}}{2}k_{0}a_{0}$
& $\frac{k_{0}^{3}}{3 \cdot 2 \cdot 1}a_{0}$ \RTS \\
\hline
4
& $\frac{k_{0}}{4}a_{3}=\frac{k_{0}}{4}\frac{k_{0}}{3} \frac{k_{0}}{2}k_{0}a_{0}$
& $\frac{k_{0}^{4}}{4 \cdot 3 \cdot 2 \cdot 1}a_{0}$ \RTS \\
\hline
\end{tabular}
\end{table}
\end{document}

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Thank you! this surely did the trick :) –  Møller Nov 27 '12 at 18:18