voorbeeld 2 uitgewerkt

$ \begin{array}{l} {\rm{Te}}\,\,{\rm{bewijzen:}}1^3 + 2^3 + ... + n^3 = \frac{1}{4}n^2 \left( {n + 1} \right)^2 \\ {\rm{Stap}}\,\,{\rm{1:}}\,\,{\rm{neem}}\,\,{\rm{n = 1}} \\ 1^3 = 1\,\,en\,\,\frac{1}{4} \cdot 1^2 \left( {1 + 1} \right)^2 = 1\,\,{\rm{Klopt}}\,\,{\rm{voor}}\,\,{\rm{n = 1}} \\ {\rm{Stap}}\,\,{\rm{2:}}\,\,{\rm{neem}}\,\,{\rm{n + 1}} \\ 1^3 + 2^3 + ... + n^3 + (n + 1)^3 = \frac{1}{4}\left( {n + 1} \right)^2 \left( {\left( {n + 1} \right) + 1} \right)^2 \\ \frac{1}{4}n^2 \left( {n + 1} \right)^2 + (n + 1)^3 = \frac{1}{4}\left( {n + 1} \right)^2 \left( {n + 2} \right)^2 \\ \frac{1}{4}\left( {n + 1} \right)^2 \left\{ {n^2 + 4\left( {n + 1} \right)} \right\} = \frac{1}{4}\left( {n + 1} \right)^2 \left( {n + 2} \right)^2 \\ \frac{1}{4}\left( {n + 1} \right)^2 \left\{ {n^2 + 4n + 4} \right\} = \frac{1}{4}\left( {n + 1} \right)^2 \left( {n + 2} \right)^2 \\ \frac{1}{4}\left( {n + 1} \right)^2 (n + 2)^2 = \frac{1}{4}\left( {n + 1} \right)^2 \left( {n + 2} \right)^2 \\ {\rm{Klopt!}} \\ \end{array} $