<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">%This command provides the text for the closed form of the sums on the page
%1 first part of the second column
\newcommand\TOneSums[1]{%
    \parbox[t]{#1}{%
      \TOneSeriesFontSize
      \begin{DisplayFormulae}{0}{0pt}{\TOneInterlineSeries}{\BigChar}{\StyleWithoutNumber}%
         \def\FmSep{\unskip\text{,}}
         \Fm{\sum_{i=1}^n i = \frac{n(n+1)}{2}} 
         \Fm{\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}}
         \def\FmSep{\relax}
         \Fm{\sum_{i=1}^n i^3 = \frac{n^2(n+1)^2}{4}}
      \end{DisplayFormulae}

      \TOneTitle{In general:}
      \AdjustSpace{-2ex plus .5ex minus .5ex}
      \begin{DisplayFormulae}{0}{0pt}{\TOneInterlineSeries}{\BigChar}{\StyleWithoutNumber}%
           %The split of this equation is tricky since it uses a variable length symbol ([)
           %depending of the size of the sums. 
           %
           %By using a rule in the first part which has the same depth and height 
           %as the sum symbol it is possible to split the equation
           %and keep the correct size of the symbol with a variable size. 
           \def\EquationPartB{\sum_{i=1}^n \left((i+1)^{m+1} - i^{m+1} - (m+1)i^m\right)}
           \settoheight{\TmpLengthA}{$\EquationPartB$} 
           \settoheight{\VSpace}{$\EquationPartB$} 
           \def\FirstPart{\sum_{i=1}^n i^m = \frac{1}{m+1} 
                                     \left[\mbox{}\rule[\VSpace]{0pt}{\TmpLengthA}\right.}
      	   \FmPartA{\FirstPart (n+1)^{m+1} - 1 -} 
      	   \FmPartB{\FirstPart}{\left.\EquationPartB\right]}
         \Fm{\sum_{i=1}^{n-1} i^m  = \frac{1}{m+1}\sum_{k=0}^m \binom{m+1}{k} B_k n^{m+1-k}} 
      \end{DisplayFormulae}

      \TOneTitle{Geometric series:}
      \begin{DisplayFormulae}{0}{0pt}{\TOneInterlineSeries}{\BigChar}{\StyleWithoutNumber}%
         \def\FmSep{\unskip\text{,}}
         \Fm{\sum_{i=0}^n c^i = \frac{1-c^{n+1}}{1-c}\MathRemark{c \neq 1}}
         \Fm{\sum_{i=0}^\infty c^i = \frac{1}{1 - c}}
         \Fm{\sum_{i=1}^\infty c^i = \frac{c}{1 - c}\MathRemark{\vert c \vert &lt; 1}}
         \Fm{\sum_{i=0}^n i c^i = \frac{nc^{n+2} - (n+1)c^{n+1} + c}{(c-1)^2}\MathRemark{c \neq 1}}
         \def\FmSep{\relax}
         \Fm{\sum_{i=0}^\infty i c^i = \frac{c}{(1 - c)^2}\MathRemark{\vert c \vert &lt; 1}}
      \end{DisplayFormulae}

      \TOneTitle{Harmonic series:}
      \begin{DisplayFormulae}{0}{0pt}{\TOneInterlineSeries}{\BigChar}{\StyleWithoutNumber}%
         \def\FmSep{\unskip\text{,}}
         \Fm{H_n = \sum_{i=1}^n \frac{1}{i}}
         \Fm{\sum_{i=1}^n iH_i = \frac{n(n+1)}{2}H_n - \frac{n(n-1)}{4}}
         \Fm{\sum_{i=1}^n H_i = (n+1)H_n - n}
         \def\FmSep{\relax}
         \Fm{\sum_{i=1}^n \binom{i}{m} H_i = \binom{n+1}{m+1} \left(H_{n+1} - \frac{1}{m+1}\right)}
      \end{DisplayFormulae}
   }%
}
</pre></body></html>