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Question
Simplify n[n! + (n – 1)!] + n2(n – 1)! + (n + 1)!
Solution
n[n! + (n – 1)!] + n2(n – 1)! + (n + 1)!
= n[n(n – 1)! + (n – 1)!] + n2(n – 1)! + (n + 1)n(n – 1)!
= (n – 1)! [n(n + 1) + n2 + (n + 1)n]
= (n – 1)! [n2 + n + n2 + n2 + n]
= (n – 1)! (3n2 + 2n)
= n(n – 1)! (3n + 2) = (3n + 2)n!
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