sum(n=1)^(99) n! (n^2 + n+1) is equal ...

`sum_(n=1)^(99) n! (n^2 + n+1) ` is equal to :

A

(100)!99-1

B

(100)(100)!-1

C

`100^100-1`

D

None of these

Verified by Experts

The correct Answer is:

B

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