%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 < 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 < 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} }% }