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If $\sum_{r=1}^{10} r!(r^3 + 6r^2 + 2r + 5) = \alpha(11!)$, then the value of $\alpha$ is equal to _______.
$\sum_{r=1}^{10} r!{(r+1)(r+2)(r+3) - 9(r+1) + 8}$
$= \sum_{r=1}^{10}{(r+3)! - (r+1)!} - 8{(r+1)! - r!}$
$= (13! + 12! - 2! - 3!) - 8(11!! - 1)$
$= (12.13 + 12 - 8) \cdot 11! - 8 + 8$
$= (160)(11!)$
Hence $\alpha = 160$
BETA
