# Command representing an expansion of a summation

I am very new to using LaTeX to write equation. And I am now learning how to create my own command. I am going to write a command \Listn to help me to do this:

\Listn{\sumup}{(#1)}{+}
$\sumup{n}$\\
\Listn{\sumupaf}{a_{#1}f(x_{#1})}{\times}
$\sumupaf{k+1}$


will give me

Then I try to create \Listn

\providecommand{\Define}[2]{\providecommand{#1}{}\renewcommand{#1}[1]{#2}}
\providecommand{\List}[3]{{#1{1}}#2{#1{2}}#2\cdots#2{#1{#3}}}
\providecommand{\Listn}[4]{\Define\f{#2}\Define{#1}{\List{\f}{#3}{#4}}}


But the fourth argument must be filled manually, like this:

\Listn{\sumup}{(#1)}{+}{#1}
$\sumup{n}$\\
\Listn{\sumupaf}{a_{#1}f(x_{#1})}{\times}{#1}
$\sumupaf{k+1}$


If I want to eliminate the fourth argument, I would like to have

\providecommand{\argument}{#1}


which is impossible.

Are there any method to overcome this?

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It would be better to state what you want to achieve and your question explicitly. At the moment people have to read both from the code, which isn't that simple. Also the title is irritating. The term 'expansion' means something different in TeX. –  Martin Scharrer Nov 26 '11 at 14:44

When you are defining a macro within another macro, the nested macro must use ##n to access its nth parameter, as #n refers to the nth parameter of the parent macro. So, you simply need to change \Listn macro to use {##1} as the last parameter allows you to eliminate the 4th parameter:

\providecommand{\Listn}[3]{\Define\f{#2}\Define{#1}{\List{\f}{#3}{##1}}}


\documentclass{article}

\providecommand{\Define}[2]{\providecommand{#1}{}\renewcommand{#1}[1]{#2}}
\providecommand{\List}[3]{{#1{1}}#2{#1{2}}#2\cdots#2{#1{#3}}}
\providecommand{\Listn}[3]{\Define\f{#2}\Define{#1}{\List{\f}{#3}{##1}}}

\begin{document}
\Listn{\sumup}{(#1)}{+}
$\sumup{n}$

\Listn{\sumupaf}{a_{#1}f(x_{#1})}{\times}
$\sumupaf{k+1}$
\end{document}

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I think you don't really need the first argument, that is \sumup or \sumupaf:

\documentclass[a4paper]{article}
\newcommand\Listn[3]{\def\next##1{#1}\next{1}#2\next{2}#2\cdots#2\next{#3}}

\begin{document}
$\Listn{(#1)}{+}{n}$

$\Listn{a_{#1}f(x_{#1})}{\times}{k+1}$
\end{document}


Here \next is a "scratch control sequence" that will be reused and redefined without problems.

The first argument to \Listn is the "structure" of the factors, the second is the operation, and the third is the last index.

If you really need to give a name to the structures for subsequent usage, you can say

\newcommand\Listn[4][\listnscratch]{%
\gdef#1##1{#2}#1{1}#3#1{2}#3\cdots#3#1{#4}}


and then $\Listn[\sumup]{(#1)}{+}{n}$ would give the same result as before, and you'll be allowed to say \sumup{3} thereon to get "(3)".

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