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

$\frac{(n+2)!-(n+1)!}{n!}$ can be simplified by common denominator n!. Therefore, $\frac{(n+2)!-(n+1)!}{n!} = (n+2)(n+1)-(n+1)$ This expression can be shown as $(n+1)(n+2-1) = (n+1)^2$ which proves that the answer is $\textbf{(D)}$ .

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2020 AMC 12B Problems/Problem 6

Problem 6

Solution