Question

If $1+\sum _{r=0}^{22}\left\{r\right(r+2)+1\}\cdot r!=k!$, then the value of $k$ is

Answer 1

$1+\sum _{r=0}^{22}\left\{r\right(r+2)+1\}\cdot r!=1+\sum _{r=0}^{22}\left\{\right(r+1-1\left)\right(r+2)+1\}r!=1+\sum _{r=0}^{22}\left[\right(r+2)!-(r+1)!]=1+\left[\begin{array}{c}2!-1!\\ 3!-2!\\ ⋮\\ 24!-23!\end{array}\right]=1+24!-1!=24!$
Therefore,
$k=24$