Opgave Toon aan: $\left( {\begin{array}{*{20}c} {n + 2} \\ k \\ \end{array}} \right) = \left( {\begin{array}{*{20}c} n \\ k \\ \end{array}} \right) + 2 \cdot \left( {\begin{array}{*{20}c} n \\ {k - 1} \\ \end{array}} \right) + \left( {\begin{array}{*{20}c} n \\ {k - 2} \\ \end{array}} \right)$ Uitwerkingen $\left( {\begin{array}{*{20}c} n \\ k \\ \end{array}} \right) + 2 \cdot \left( {\begin{array}{*{20}c} n \\ {k - 1} \\ \end{array}} \right) + \left( {\begin{array}{*{20}c} n \\ {k - 2} \\ \end{array}} \right) =$ \eqalign{\frac{{n!}}{{k!\left( {n - k} \right)!}} + 2\frac{{n!}}{{\left( {k - 1} \right)!\left( {n - k + 1} \right)!}} + \frac{{n!}}{{(k - 2)!\left( {n - k + 2} \right)!}} =} \eqalign{ (n - k + 1)(n - k + 2)\frac{{n!}}{{k!\left( {n - k + 2} \right)!}} +... } \eqalign{ \,\,\,\,\,\,... + 2k\left( {n - k + 2} \right)\frac{{n!}}{{k!\left( {n - k + 2} \right)!}} + ... } \eqalign{ \,\,\,\,\,\,\,\,\,\,\,\,... + k(k - 1)\frac{{n!}}{{k!\left( {n - k + 2} \right)!}} = } \eqalign{ \frac{{n!}}{{k!\,\left( {n - k + 2} \right)!}}\left( {(n - k + 1)(n - k + 2) + 2k\left( {n - k + 2} \right) + k(k - 1)} \right) = } \eqalign{ \frac{{n!}}{{k!\,\left( {n - k + 2} \right)!}}\left( {(n + 1)(n + 2)} \right) = } \eqalign{ \frac{{n! \cdot (n + 1)(n + 2)}}{{k!\left( {n - k + 2} \right)!}} = } \eqalign{ \frac{{(n + 2)!}}{{k!\left( {n - k + 2} \right)!}} = } \eqalign{ \left( {\begin{array}{*{20}c} {n + 2} \\ k \\ \end{array}} \right) } Toelichting Je wilt aantonen dat: $\left( {\begin{array}{*{20}c} {n + 2} \\ k \\ \end{array}} \right) = \left( {\begin{array}{*{20}c} n \\ k \\ \end{array}} \right) + 2 \cdot \left( {\begin{array}{*{20}c} n \\ {k - 1} \\ \end{array}} \right) + \left( {\begin{array}{*{20}c} n \\ {k - 2} \\ \end{array}} \right)$ Als je die termen uitschrijft krijg je: \eqalign{\frac{{(n + 2)!}}{{k!\left( {n - k + 2} \right)!}} = \frac{{n!}}{{k!\left( {n - k} \right)!}} + 2\frac{{n!}}{{\left( {k - 1} \right)!\left( {n - k + 1} \right)!}} + \frac{{n!}}{{(k - 2)!\left( {n - k + 2} \right)!}}} De kunst is nu om alle termen aan de rechter kant te schrijven als: \eqalign{\frac{{...}}{{k!\left( {n - k + 2} \right)!}}} Het is (als het ware) gelijknamig maken! Daarna kan je de termen rechts optellen en kijken wat er bij de teller moet staan... Begrip en inzicht Je zou ook in de driehoek van Pascal kunnen kijken: $\begin{array}{l} \left( {\begin{array}{*{20}c} n \\ k \\ \end{array}} \right) + 2 \cdot \left( {\begin{array}{*{20}c} n \\ {k - 1} \\ \end{array}} \right) + \left( {\begin{array}{*{20}c} n \\ {k - 2} \\ \end{array}} \right) = \\ \left( {\begin{array}{*{20}c} n \\ {k - 2} \\ \end{array}} \right) + \left( {\begin{array}{*{20}c} n \\ {k - 1} \\ \end{array}} \right) + \left( {\begin{array}{*{20}c} n \\ {k - 1} \\ \end{array}} \right) + \left( {\begin{array}{*{20}c} n \\ k \\ \end{array}} \right) = \\ \left( {\begin{array}{*{20}c} {n + 1} \\ {k - 1} \\ \end{array}} \right) + \left( {\begin{array}{*{20}c} {n + 1} \\ k \\ \end{array}} \right) = \\ \left( {\begin{array}{*{20}c} {n + 2} \\ k \\ \end{array}} \right) \\ \end{array}$ ...dan is het minder ingewikkeld dan je denkt.  Gebruik daarbij de eigenschap: $\left( {\begin{array}{*{20}c} n \\ k \\ \end{array}} \right) = \left( {\begin{array}{*{20}c} {n - 1} \\ {k - 1} \\ \end{array}} \right) + \left( {\begin{array}{*{20}c} {n - 1} \\ k \\ \end{array}} \right)$