2020 AMC 12B Problems/Problem 6

Problem 6

For all integers $n \geq 9,$ the value of \[\frac{(n+2)!-(n+1)!}{n!}\]is always which of the following?

$\textbf{(A) } \text{a multiple of }4 \qquad \textbf{(B) } \text{a multiple of }10 \qquad \textbf{(C) } \text{a prime number} \\ \textbf{(D) } \text{a perfect square} \qquad \textbf{(E) } \text{a perfect cube}$

Solution

We first expand the expression: \[\frac{(n+2)!-(n+1)!}{n!} = \frac{(n+2)(n+1)n!-(n+1)n!}{n!}\]

We can now divide out a common factor of $n!$ from each term of this expression:

\[(n+2)(n+1)-(n+1)\]

Factoring out $(n+1)$, we get \[(n+1)(n+2-1) = (n+1)^2\]

which proves that the answer is $\boxed{\textbf{(D)} \text{ a perfect square}}$.

Solution 2

Factor out an $n!$ to get: $\frac{(n+2)!-(n+1)!}{n!} = (n+2)(n+1)-(n+1)$ Now, without loss of generality, test values of $n$ until only one answer choice is left valid:

$n = 1 \implies (3)(2) - (2) = 4$, knocking out $\textbf{B}$, $\textbf{C}$, and $\textbf{E}$. \[\] $n = 2 \implies (4)(3) - (3) = 9$, knocking out $\textbf{A}$.

This leaves $\boxed{\textbf{(D)} \text{ a perfect square}}$ as the only answer choice left.

With further testing it becomes clear that for all $n$, $(n+2)(n+1)-(n+1) = (n+1)^{2}$, proved in Solution 1.

~DBlack2021

Video Solution

https://youtu.be/6ujfjGLzVoE

~IceMatrix

See Also

2020 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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