Find the sum of the series $(1^{2} + 1) 1! + (2^{2} + 1) 2! + (3^{2} + 1) 3! + . . . . . . (n^{2} + 1) n!$

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Solution

$n! = 1 \times 2 \dots n$
$(n + 1) n! = 1 \times 2 \dots n \times (n + 1) = (n + 1)!$ $(P)$
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$S = n \sum k = 1 (k^{2} + 1) k!$
$= n \sum k = 1 (k^{2} + 2 k + 1 - 2 k) k!$
$= n \sum k = 1 ((k + 1)^{2} - 2 k) k!$
$= n \sum k = 1 (k + 1)^{2} k! - 2 k \cdot k!$
$= n \sum k = 1 (k + 1) \cdot (k + 1)! - 2 k \cdot k!$ (using $(P)$ )
$= n \sum k = 1 [(k + 1) \cdot (k + 1)! - k \cdot k!] - n \sum k = 1 k \cdot k!$
$= (n + 1) (n + 1)! - 1 \cdot 1! - n \sum k = 1 (k + 1 - 1) k!$
$= (n + 1) (n + 1)! - 1 - [n \sum k = 1 (k + 1) k! - k!]$
$= (n + 1) (n + 1)! - 1 - [n \sum k = 1 (k + 1)! - k!]$ (using $(P)$ )
$= (n + 1) (n + 1)! - 1 - [(n + 1)! - 1!]$
$= (n + 1) (n + 1)! - (n + 1)!$
$= n (n + 1)!$

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