Question:
If $\sum_{r=1}^{10} r !\left(r^{3}+6 r^{2}+2 r+5\right)=\alpha(11 !)$, then the
value of $\alpha$ is equal to_______
Solution:
$\sum_{r=1}^{10} r !\{(r+1)(r+2)(r+3)-9(r+1)+8\}$
$=\sum_{\mathrm{r}=1}^{10}[\{(\mathrm{r}+3) !-(\mathrm{r}+1) !\}-8\{(\mathrm{r}+1) !-\mathrm{r} !\}]$
$=(13 !+12 !-2 !-3 !)-8(11 !-1)$
$=(12.13+12-8) \cdot 11 !-8+8$
$=(160)(11) !$
Hence $\alpha=160$