bewijs van de regel van Pascal

 In de driehoek van Pascal is elk getal de som van de twee getallen die er schuin boven staan. $\left( {\begin{array}{*{20}c} n\\ {k-1}\\ \end{array}} \right) + \left( {\begin{array}{*{20}c} n\\ k\\ \end{array}} \right) = \left( {\begin{array}{*{20}c} {n+1}\\ k\\ \end{array}} \right)$ Bewijs $\begin{gathered} \left. \begin{gathered} \left( {\begin{array}{*{20}c} n \\ k \\ \end{array} } \right) = \frac{{n!}} {{k!\, \cdot \left( {n - k} \right)!}}\, \\ \,\left( {\begin{array}{*{20}c} n \\ {k - 1} \\ \end{array} } \right) = \frac{{n!}} {{(k - 1)!\, \cdot \left( {n - (k - 1)} \right)!}} \\ \end{gathered} \right\} \to \left( {\begin{array}{*{20}c} {n + 1} \\ k \\ \end{array} } \right) = \frac{{(n + 1)!}} {{k!\, \cdot (n + 1 - k)!}} \\ \left( {\begin{array}{*{20}c} n \\ {k - 1} \\ \end{array} } \right) + \left( {\begin{array}{*{20}c} n \\ k \\ \end{array} } \right) = \frac{{n!}} {{(k - 1)!\, \cdot \left( {n - (k - 1)} \right)!}} + \frac{{n!}} {{k!\, \cdot \left( {n - k} \right)!}} \\ \left( {\begin{array}{*{20}c} n \\ {k - 1} \\ \end{array} } \right) + \left( {\begin{array}{*{20}c} n \\ k \\ \end{array} } \right) = \frac{{n!}} {{(k - 1)!\, \cdot \left( {n - k + 1} \right)!}} \cdot \frac{k} {k} + \frac{{n!}} {{k!\, \cdot \left( {n - k} \right)!}} \cdot \frac{{n - k + 1}} {{n - k + 1}} \\ \left( {\begin{array}{*{20}c} n \\ {k - 1} \\ \end{array} } \right) + \left( {\begin{array}{*{20}c} n \\ k \\ \end{array} } \right) = \frac{{n!\, \cdot k}} {{k!\, \cdot \left( {n - k + 1} \right)!}} + \frac{{n!\left( {n - k + 1} \right)}} {{k!\, \cdot \left( {n - k + 1} \right)!}} \\ \left( {\begin{array}{*{20}c} n \\ {k - 1} \\ \end{array} } \right) + \left( {\begin{array}{*{20}c} n \\ k \\ \end{array} } \right) = \frac{{n!\, \cdot k + n!\left( {n - k + 1} \right)}} {{k!\, \cdot \left( {n - k + 1} \right)!}} \\ \left( {\begin{array}{*{20}c} n \\ {k - 1} \\ \end{array} } \right) + \left( {\begin{array}{*{20}c} n \\ k \\ \end{array} } \right) = \frac{{n!\left( {k + n - k + 1} \right)}} {{k!\, \cdot \left( {n - k + 1} \right)!}} \\ \left( {\begin{array}{*{20}c} n \\ {k - 1} \\ \end{array} } \right) + \left( {\begin{array}{*{20}c} n \\ k \\ \end{array} } \right) = \frac{{n!\left( {n + 1} \right)}} {{k!\, \cdot \left( {n - k + 1} \right)!}} \\ \left( {\begin{array}{*{20}c} n \\ {k - 1} \\ \end{array} } \right) + \left( {\begin{array}{*{20}c} n \\ k \\ \end{array} } \right) = \frac{{\left( {n + 1} \right)!}} {{k!\, \cdot \left( {n - k + 1} \right)!}} \\ \left( {\begin{array}{*{20}c} n \\ {k - 1} \\ \end{array} } \right) + \left( {\begin{array}{*{20}c} n \\ k \\ \end{array} } \right) = \left( {\begin{array}{*{20}c} {n + 1} \\ k \\ \end{array} } \right) \\ \end{gathered}$