voorbeeld 4 uitgewerkt

$\begin{array}{l} {\rm{Te}}\,\,{\rm{bewijzen:}}1^2 + 2^2 + ... + n^2 = \frac{1}{6}\,n\left( {n + 1} \right)\left( {2n + 1} \right) \\ {\rm{Stap}}\,{\rm{1:}} \\ {\rm{Neem}}\,\,{\rm{n = 1:}} \\ \left. \begin{array}{l} 1^2 = 1 \\ \frac{1}{6}\,n\left( {n + 1} \right)\left( {2n + 1} \right) = \frac{1}{6} \cdot \,1\left( {1 + 1} \right)\left( {2 \cdot 1 + 1} \right) = \frac{1}{6} \cdot \,2 \cdot 3 = 1 \\ \end{array} \right\} \Rightarrow {\rm{klopt!}} \\ {\rm{Neem}}\,\,\,{\rm{n + 1:}} \\ 1^2 + 2^2 + ... + n^2 + \left( {n + 1} \right)^2 = \frac{1}{6}\,\left( {n + 1} \right)\left( {n + 2} \right)\left( {2n + 3} \right) \\ \frac{1}{6}\,n\left( {n + 1} \right)\left( {2n + 1} \right) + \left( {n + 1} \right)^2 = \,\frac{1}{6}\left( {n + 1} \right)\left( {n + 2} \right)\left( {2n + 3} \right) \\ \frac{1}{6}\,n\left( {2n + 1} \right) + n + 1 = \,\frac{1}{6}\left( {n + 2} \right)\left( {2n + 3} \right) \\ \,n\left( {2n + 1} \right) + 6n + 6 = \,\left( {n + 2} \right)\left( {2n + 3} \right) \\ 2n^2 + 7n + 6 = 2n^2 + 7n + 6 \\ {\rm{Klopt!}} \\ \end{array}$