If $\sum\limits_{r = 1}^{10} {r!({r^3} + 6{r^2} + 2r + 5) = \alpha (11!)}$, then the value of $\alpha$ is equal to ___________.
Answer (integer)
160
Solution
<p>$\sum\limits_{r = 1}^{10} {r![(r + 1)(r + 2)(r + 3) - 9(r + 1) + 8]}$</p>
<p>$= \sum\limits_{r = 1}^{10} {[\{ (r + 3)! - (r + 1)!\} - 8\{ (r + 1)! - r!\} ]}$</p>
<p>$= (13! + 12! - 2! - 3!) - 8(11! - 1)$</p>
<p>$= (12\,.\,13 + 12 - 8)\,.\,11! - 8 + 8 = (160)(11!)$</p>
<p>Therefore, $\alpha = 160$</p>
About this question
Subject: Mathematics · Chapter: Permutations and Combinations · Topic: Fundamental Counting Principle
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