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Difference between revisions of "2020 AMC 12B Problems/Problem 6"

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This expression can be shown as <cmath>\(n+1)(n+2-1) = (n+1)^2</cmath>
 
This expression can be shown as <cmath>\(n+1)(n+2-1) = (n+1)^2</cmath>
 
which proves that the answer is <math>\textbf{(D)}</math>.
 
which proves that the answer is <math>\textbf{(D)}</math>.
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==Problem 6==
 
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For all integers <math>n \geq 9,</math> the value of
 
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<cmath>\frac{(n+2)!-(n+1)!}{n!}</cmath>is always which of the following?
 
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<math>\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}</math>
 

Revision as of 20:25, 7 February 2020

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\] (Error compiling LaTeX. Unknown error_msg)

which proves that the answer is $\textbf{(D)}$.

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