For all integers n ≥ 9, the value of ((n + 2)! - (n + 1)!)/(n!) is always which of the following?

2020 AMC 12B Problem 6

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For all integers

n \ge 9,

the value of

\frac{(n + 2)! - (n + 1)!}{n!}

is always which of the following?

a multiple of

4

a multiple of

10

a prime number

a perfect square

a perfect cube

Factor

(n + 1)!

out of the numerator

(n + 2)! - (n + 1)!

= (n + 1)!\,\big[(n + 2) - 1\big]

= (n + 1)!\,(n + 1)

Factor

(n + 1)!

from the numerator:

\begin{gathered} (n + 2)! - (n + 1)! \\ {}= (n + 1)!\,\big[(n + 2) - 1\big] \\ {}= (n + 1)!\,(n + 1). \end{gathered}

Dividing by

n!

leaves

\dfrac{(n + 1)!\,(n + 1)}{n!}

= (n + 1)(n + 1)

= (n + 1)^2,

which is always a perfect square.

Thus, the correct answer is D.