1^ste manier : met de regel van SARRUS
\(\left|\begin{matrix}1&n&n(n+1)\\1&n+1&(n+1)(n+2)\\1&n+2&(n+2)(n+3)\\\end{matrix}\right|=(n+1)(n+2)(n+3)+n(n+1)(n+2)\\+n(n+1)(n+2)-n(n+1)-(n+1)(n+2)^2-n(n+2)(n+3)\\\small=(n+1)(n+2)(n+3)+2n(n+1)(n+2)-n(n+1)^2-(n+1)(n+2)^2-n(n+2)(n+3)\\=(n+2)(n+3)+2n(n+1)(n+2)-n(n+1)^2-(n+1)(n+2)^2\\=n^2+5n+6+2n(n+3n+2)-n(n+2n+1)-(n+1)(n^2+4n+4)\\=n^2+5n+6+2n^3+6n^2+4n-n^3-2n^2-n-n^3-4n^2-4n-n^2-4n-4\\=6-4=2\)
2^de manier : door toepassing van een belangrijke rekenregel bij determinanten
      (eerst 2^de rij van de 3^de aftrekken, daarna 1^ste van 2^de)
\(\left|\begin{matrix}1&n&n(n+1)\\1&n+1&(n+1)(n+2)\\1&n+2&(n+2)(n+3)\\\\\end{matrix}\right|\overset{R3-R2}=\left|\begin{matrix}1&n&n(n+1)\\1&n+1&(n+1)(n+2)\\0&1&2(n+2)\\\end{matrix}\right|\\\overset{R2-R1}=\left|\begin{matrix}1&n&n(n+1)\\0&1&2(n+1)\\0&1&2(n+2)\\\end{matrix}\right|=2\left|\begin{matrix}1&n+1\\1&n+2\\\end{matrix}\right|=2.1=2\)
3^de manier : de mooiste manier (gevonden door een van mijn leerlingen)
Hierbij gaan we op zoek naar de determinant van VANDERMONDE
die zegt dat \(\small\begin{vmatrix}1&a&a^2\\1&b&b^2\\1&c&c^2\\\end{vmatrix}=(a-b)(b-c)(c-a)\)
\(\left|\begin{matrix}1&n&n(n+1)\\1&n+1&(n+1)(n+2)\\1&n+2&(n+2)(n+3)\\\end{matrix}\right|\overset{K3-K2}=\left|\begin{matrix}1&n&n(n+1)-n\\1&n+1&(n+1)(n+2)-(n+1)\\0&n+2&(n+2)(n+3)-(n+2)\\\end{matrix}\right|\\\begin{vmatrix}1&n&n^2\\1&n+1&(n+1)^2\\0&n+2&(n+2)^2\\\end{vmatrix}\overset{Vandermonde}=[n-(n+1)][(n+1)-(n+2)][(n+2)-n]\\=(-1)(-1).2=2\)