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

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m (Punctuated Sol 1 accordingly.)
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==Solution 1==
 
==Solution 1==
 
We first expand the expression:
 
We first expand the expression:
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<cmath>\frac{(n+2)!-(n+1)!}{n!} = \frac{(n+2)(n+1)n!-(n+1)n!}{n!}</cmath>
+
<cmath>\frac{(n+2)!-(n+1)!}{n!} = \frac{(n+2)(n+1)n!-(n+1)n!}{n!}.</cmath>
−
 
 
 
We can now divide out a common factor of <math>n!</math> from each term of this expression:
 
We can now divide out a common factor of <math>n!</math> from each term of this expression:
−
 
+
<cmath>(n+2)(n+1)-(n+1).</cmath>
−
<cmath>(n+2)(n+1)-(n+1)</cmath>
+
Factoring out <math>(n+1),</math> we get <cmath>(n+1)(n+2-1) = (n+1)^2,</cmath>
−
 
 
−
Factoring out <math>(n+1),</math> we get <cmath>(n+1)(n+2-1) = (n+1)^2</cmath>
 
−
 
 
 
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>
  

Revision as of 10:21, 18 October 2021

Problem

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 1

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

In the numerator, we 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.

This solution does not consider the condition $n \geq 9.$ The reason is that, with further testing it becomes clear that for all $n,$ we get \[(n+2)(n+1)-(n+1) = (n+1)^{2},\] as proved in Solution 1. The condition $n \geq 9$ was most likely added to encourage the selection of choice $\textbf{(B)}.$

~DBlack2021 (Solution Writing)

~Countmath1 (Minor edits in formatting)

~MRENTHUSIASM (Edits in Logic)

Video Solution

https://youtu.be/ba6w1OhXqOQ?t=2234

~ pi_is_3.14

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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