First slide

Permutations

Question

Sum of the series $\sum_{r = 1}^{n} (r^{2} + 1) (r!)$ is

Moderate

A

$(n + 1)!$
B

$(n + 2)! - 1$
C

$n (n + 1)!$
D

none of these

Solution

We can write
$r^{2} + 1 = (r + 2) (r + 1) - 3 (r + 1) + 2$
Thus $\sum_{r = 1}^{n} (r^{2} + 1) (r!) = \sum_{r = 1}^{n} [(r + 2) (r + 1) - (r + 1) - 2 {(r + 1) - 1}] r!$
$\begin{array}{l} = \sum_{r = 1}^{n} [(r + 2)! - (r + 1)!] - 2 \sum_{r = 1}^{n} {(r + 1)! - r!} \\ = (n + 2)! - 2! - 2 {(n + 1)! - 1} = n (n + 1)! \end{array}$

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