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

2020 AMC 12B Problem 6 — Solution

2020 AMC 12B Problem 6

Attempt Problem 6 of the 2020 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2020 AMC 12B solutions, or check the answer key.

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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

Answer: D

Concepts:factorial factoring perfect square

Difficulty rating: 1270

Solution:

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.

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