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

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(Solution 2)
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which proves that the answer is <math>\boxed{\textbf{(D)} \text{ a perfect square}}</math>.
 
which proves that the answer is <math>\boxed{\textbf{(D)} \text{ a perfect square}}</math>.
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==Solution 2==
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Factor out an <math>n!</math> to get: <math>\frac{(n+2)!-(n+1)!}{n!} = (n+2)(n+1)-(n+1)</math> Now, without loss of generality, test values of <math>n</math> until only one answer choice is left valid:
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<math>n = 1 \implies (3)(2) - (2) = 4</math>, knocking out <math>\textbf{B}</math>, <math>\textbf{C}</math>, and <math>\textbf{E}</math>.
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<cmath> </cmath>
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<math>n = 2 \implies (4)(3) - (3) = 9</math>, knocking out <math>\textbf{A}</math>.
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This leaves <math>\boxed{\textbf{(D)} \text{ a perfect square}}</math> as the only answer choice left.
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With further testing it becomes clear that for all <math>n</math>, <math>(n+2)(n+1)-(n+1) = (n+1)^{2}</math>, proved in Solution 1.
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~DBlack2021
  
 
==Video Solution==
 
==Video Solution==

Revision as of 07:32, 10 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

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 (Problems • Answer Key • Resources)
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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