nice Factorial equation exercise asked by Dan Situra in many math groups

Exercise:

Solve in $\mathbb{N}$ , $\frac{\sum\nolimits_{k=1}^{n}{{{k}^{2}}\left( k+1 \right)}!-2}{\left( n+1 \right)!}=108$

Solution: we have ${{k}^{2}}\left( k+1 \right)!=\left( {{k}^{2}}-4+4 \right)\left( k+1 \right)!=\left( {{k}^{2}}-4 \right)\left( k+1 \right)!+4\left( k+1 \right)!$

$=\left( k+2 \right)\left( k-2 \right)\left( k+1 \right)!+4\left( k+1 \right)!=\left( k-5+3 \right)\left( k+2 \right)\left( k+1 \right)!+4\left( k+1 \right)!$

$=\left( k+3 \right)\left( k+2 \right)\left( k+1 \right)!\,\,-5\left( k+2 \right)\left( k+1 \right)!+4\left( k+1 \right)!$

$\Leftrightarrow {{k}^{2}}\left( k+1 \right)!=\left( k+3 \right)!-5\left( k+2 \right)!+4\left( k+1 \right)!=\left( k+3 \right)!-4\left( k+2 \right)!-\left( k+2 \right)!+4\left( k+1 \right)!$

                 $=\left[ \left( k+3 \right)!-\left( k+2 \right)! \right]-4\left[ \left( k+2 \right)!-\left( k+1 \right)! \right]$

So $\sum\nolimits_{k=1}^{n}{{{k}^{2}}\left( k+1 \right)!=\sum\nolimits_{k=1}^{n}{\left( \left( k+3 \right)!-\left( k+2 \right)! \right)-4\sum\nolimits_{k=1}^{n}{\left( \left( k+2 \right)!-\left( k+1 \right)! \right)}}}$

but $\sum\nolimits_{k=1}^{n}{\left( \left( k+3 \right)!-\left( k+2 \right)! \right)=4!-3!+5!-4!+6!-5!+....+\left( n+2 \right)!+\left( n+3 \right)!-\left( n+2 \right)!}$

                                       $=\left( n+3 \right)!-3!$

$\sum\nolimits_{k=1}^{n}{\left( \left( k+2 \right)!-\left( k+1 \right)! \right)=3!-2!+4!-3!+5!-4!+...+\left( n+1 \right)!+\left( n+2 \right)!-\left( n-1 \right)!}$

                                  $=\left( n+2 \right)!-2!$

So \(\sum\nolimits_{k=1}^{n}{{{k}^{2}}\left( k+1 \right)!=\left( n+3 \right)!-4\left( n+2 \right)!+8-6=\left( n+3 \right)!-4\left( n+2 \right)!+2!}\)

                          $=\left( n+3 \right)\left( n+2 \right)\left( n+1 \right)!-4\left( n+2 \right)\left( n+1 \right)!+2$

                           $=\left( n+1 \right)!\left( \left( n+3 \right)\left( n+2 \right)-4\left( n+2 \right) \right)+2$

Thus $\frac{\sum\nolimits_{k=1}^{n}{{{k}^{2}}\left( k+1 \right)!-2}}{\left( n+1 \right)!}=\left( n+3 \right)\left( n+2 \right)-4\left( n+2 \right)=108$

So $\left( n+2 \right)\left( n+3-4 \right)=108\Leftrightarrow \left( n+2 \right)\left( n-1 \right)=108$

$\Leftrightarrow {{n}^{2}}+n-110=0\Leftrightarrow {{n}^{2}}+n+\frac{1}{4}-\frac{1}{4}-110=0\Leftrightarrow {{\left( n+\frac{1}{2} \right)}^{2}}=\frac{441}{4}\Leftrightarrow n=\frac{-1\pm \sqrt{441}}{2}$

$n=\frac{20}{2}=10\,\,\,or\,\,n=\frac{-22}{2}=-11$ (rejected ) thus the accepted root in $\mathbb{N}$ is 10