# Prove the following

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:

$\sum_{r=1}^{10} \mathrm{r} !\{(\mathrm{r}+1)(\mathrm{r}+2)(\mathrm{r}+3)-9(\mathrm{r}+1)+8\}$

$=\sum_{\mathrm{r}=1}^{10}[\{(\mathrm{r}+3) !-(\mathrm{r}+1) !\}-8\{(\mathrm{r}+1) !-\mathrm{r} !\}]$

$=(13 !+12 !-2 !-3 !)-8(11 !-1)$

$=(12.13+12-8) \cdot 11 !-8+8$

$=(160)(11) !$

Hence $\alpha=160$