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

(n + 1) . (n + 2)!

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n . (n + 1)!

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(n + 1) . (n + 1)!

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None of the above

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Solution

The correct option is B $n . (n + 1)!$
The given equation can be written as
$\sum (n^{2} + 1) . n!$
$\sum [n . (n + 1) - (n - 1)] . n!$
$\sum n . (n + 1)! - (n - 1) . n!$
Expanding the above equation we get
$(1.2! - 0) + (2.3! - 1.2!) + (3.4! - 2.3!) + . . . . . . . (n - 1) . n! - (n - 2) . (n - 1)! + n . (n + 1)! - (n - 1) . n!$
By inspection of above equation we can see that all term cancelled out except $n . (n + 1)!$
So correct answer is B

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