The sum of $(1^{2} - 1 + 1) (1!) + (2^{2} - 2 + 1) (2!) + . . . . + (n^{2} - n + 1) (n!)$ is.

(n + 2)!

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(n - 1) (n + 1)! + 1

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(n + 2)! - 1

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n (n + 1)!) - 1

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Solution

The correct option is B $(n - 1) (n + 1)! + 1$
its $[n^{2} - n + 1] * n!$
which is $[n^{2} + 3 n + 2 - 4 n - 4 + 3] n!$
which is $[n + 2]! - 4 [n + 1]! + 3 n!$
sum from $1 t o n$
to get $[n + 1]! + [n + 2]! - 4 [n + 1]! - 4 * 2! + 3 [1! + 2!]$
= $[n + 1]! \times [n + 2 - 3] + 1$
$B$ is correct

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