%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}