Sum of (example: 0+1+2+3 = 6 , 0+1+2+3+4+5+6+7 = 28 and so on)

Question

Can anyone help please? I need to know how to make the sum of value S(n) = 0+1+2+3+4...+(n-1)+n (example: 0+1+2+3 = 6 , 0+1+2+3+4+5+6+7 = 28 and so on) on Latex

Answer 1

The difficult task is generating the terms of the sequence, not computing the sum, of course; I present a macro that prints all the terms or just the sum. You can define a different starting point and another difference (defaults 0 and 1).

\documentclass{article}
\usepackage{xparse}

\ExplSyntaxOn
\NewDocumentCommand{\arithmeticsequence}{sO{}m}
 {
  \group_begin:
  \keys_set:nn {rudstep/arseq} { #2 }
  \IfBooleanTF{#1}
   { \rudstep_arseq_sum:n { #3 } }
   { \rudstep_arseq_full:n { #3 } }
  \group_end:
 }

\keys_define:nn { rudstep/arseq }
 {
  diff  .int_set:N = \l_rudstep_diff_int,
  start .int_set:N = \l_rudstep_start_int,
  diff  .initial:n = 1,
  start .initial:n = 0,
 }
\seq_new:N \l_rudstep_terms_seq

\cs_new_protected:Npn \rudstep_arseq_full:n #1
 {
  \seq_clear:N \l_rudstep_terms_seq
  \int_step_inline:nnnn
   { \l_rudstep_start_int } % start
   { \l_rudstep_diff_int }  % step
   { \l_rudstep_start_int + #1*\l_rudstep_diff_int }  % end
   { \seq_put_right:Nn \l_rudstep_terms_seq { ##1 } }
  $\seq_use:Nn \l_rudstep_terms_seq { + } = \rudstep_arseq_sum:n { #1 }$
 }
\cs_new:Npn \rudstep_arseq_sum:n #1
 {
  \int_eval:n { (#1+1)*(2*\l_rudstep_start_int+#1*\l_rudstep_diff_int)/2 }
 }
\ExplSyntaxOff

\begin{document}
Just the sum of five terms after 0: \arithmeticsequence*{5}

The whole sequence: \arithmeticsequence{3}

Or: \arithmeticsequence{4}

Start from 1: \arithmeticsequence[start=1]{3}

\bigskip

A big example: 
\arithmeticsequence[start=81297,diff=198]{180}

\end{document}

Of course, if you know how to do it in Lisp, then here's the way:

\documentclass{article}
\usepackage{lisp-on-tex}

\begin{document}

\lispinterp{
  (\define \intsum
    (\lambda (\n)
      (\lispif (\= \n :0 ) :0
        (\+ (\intsum (\- \n :1)) \n))))}

\lispinterp{(\texprint(\intsum:100))}

\end{document}

that prints, as is well known,

5050

Note: shamelessly adapted from the documentation of lisp-on-tex.

The e-TeX version of the same idea, using recursion, can be

\documentclass{article}

\newcommand{\Sn}[1]{%
  \number\numexpr #1
    \ifnum\numexpr#1-1>0
      +\expandafter\Sn\expandafter{\number\numexpr#1-1}
    \fi
  \relax
}

\begin{document}

\Sn{10}

\Sn{100}

\end{document}

Note that this is fully expandable. The macro \Sn starts from the argument and if it's bigger than 1 asks to expand \Sn with the argument decreased by 1. One might start also from 0, but I'll leave it as an exercise to the interested reader.

Answer 2

\documentclass{article}

\newcommand\foo[1]{\the\numexpr((#1)*(#1+1))/2\relax}

% or if you want to print the terms
\newcommand\foob[1]{$\fooc{#1}{0}=\foo{#1}$}
\newcommand\fooc[2]{\the\numexpr#2\relax\ifnum#1=#2\relax\else+\fooc{#1}{\numexpr#2+1\relax}\fi}

\begin{document}

\foo{8} and \foo{5}

\foob{8} and \foob{5}

\end{document}

Answer 3

Using Lua is probably a huge overkill in this situation, but it shows off how one can easily integrate Lua in LaTeX.

The code might also be easier to grasp for programmers who are beginners in LaTeX ;)

% !TEX TS-program = lualatex
% !TEX encoding = UTF-8 Unicode
\documentclass{article}

\usepackage[utf8]{luainputenc}
\usepackage{luacode}

% The code won't work as-is if you move it into \directlua{}.
% If you do so, be sure to replace "~" (in i ~= startInt) by \string~
\begin{luacode}
    function calcSum(startInt, endInt)
        local sum = 0
        for i=startInt, endInt do
            sum = sum + i
        end
        return sum
    end

    function writeSum(startInt, endInt)
        local sum = 0
        local tex = ""
        for i=startInt, endInt do
            if i ~= startInt then
                tex = tex .. " + "
            end
            sum = sum + i
            tex = tex .. i
        end
        tex = tex .. " = " .. sum
        return tex
    end
\end{luacode}

\def\calcSum#1#2{\directlua{
    tex.print(calcSum(tonumber(#1), tonumber(#2)))
}}
\def\writeSum#1#2{\directlua{
    tex.print(writeSum(tonumber(#1), tonumber(#2)))
}}

\begin{document}
    \noindent
    The sum of the numbers \(0, 1, 2, \ldots, 28\) is \(\calcSum{0}{28}\).\\
    Another sum: \(\writeSum{100}{103}\).
\end{document}

Answer 4

Here is a short expl3 approach using \int_step_inline:nnnn. This carries out the explicit sum and doesn't rely on Gauss' reduction. So this is for illustration purposes only.

\documentclass{article}
\usepackage{xparse}
\ExplSyntaxOn
\cs_generate_variant:Nn \int_to_arabic:n { V }

\cs_new:Npn \rudstep_sum:n #1
 {
  \int_zero:N \l_tmpa_int
  \int_step_inline:nnnn { 1 } { 1 } { #1 }
   {
    \int_add:Nn \l_tmpa_int { ##1 }
   }
  \int_to_arabic:V \l_tmpa_int
 }

\NewDocumentCommand \calcsum {m}
 {
  \rudstep_sum:n { #1 }
 }
\ExplSyntaxOff
\begin{document}
\calcsum{8}
\end{document}

Answer 5

Using my package calculator, you can perform arithmetic calculations comfortably. This code solves your problem:

\documentclass{article}
\usepackage{ifthen}
\usepackage{calculator}

  \newcounter{n}
  \newcommand{\sumZeroToN}[2]{%
      \COPY{0}{#2}
      \whiledo{\not{\value{n}>#1}}{%
          \ADD{#2}{\value{n}}{#2}\stepcounter{n}}}

\begin{document}
    \sumZeroToN{100}{\sol}
    \[
         \sum_{n=0}^{100} n = \sol      
    \]
\end{document}

Answer 6

An expandable solution using the package bnumexpr for computations and \xintiloop for expandably looping with an index.

\documentclass{article}
\usepackage{bnumexpr}% minimally extends \numexpr to big integers (new on CTAN
                     % as of 2014/09/22) 

% for \xintiloop: (for fun)
\usepackage{xinttools}% automatically loaded by xint, which is loaded by
                      % bnumexpr, but in the future bnumexpr will only load a
                      % smaller part of xint, and xint itself will not load
                      % xinttools anymore.

\newcommand{\ArithmeticSequenceAndItsSum}[3]{% 
  % #1= initial term, A
  % #2= common difference, d
  % #3= number of terms, N
  % sum is A+(A+d)+...+(A+(N-1)d)=N*A+d*N(N-1)/2 
  % (cf. Gauss Werke, Kindergarten Abteilung)
  %
  #1% assume N at least one and that #1 does not need to be evaluated
  \xintiloop [1+1]
  \unless\ifnum#3=\xintiloopindex
  +\thebnumexpr#1+\xintiloopindex*#2\relax
  \repeat
  =\thebnumexpr #3*#1+#2*#3*(#3-1)/2\relax
}

\begin{document}

$\ArithmeticSequenceAndItsSum {0}{1}{10}$

$\ArithmeticSequenceAndItsSum {1}{2}{10}$

\noindent $\ArithmeticSequenceAndItsSum {111111111}{111111111}{100}$

And now go see the log for some bigger integers!!

\message{\ArithmeticSequenceAndItsSum {123456789}{987654321}{1000}}

\end{document}