Difference between revisions of "2020 AMC 12B Problems/Problem 6"
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==Solution== | ==Solution== | ||
| − | <cmath>\frac{(n+2)!-(n+1)!}{n!}</cmath> can | + | We first expand the expression: |
| − | + | <cmath>\frac{(n+2)!-(n+1)!}{n!} = \frac{(n+2)(n+1)n!-(n+1)n!}{n!}</cmath> | |
| − | This | + | |
| − | + | 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> | ||
| + | |||
| + | 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>. | ||
| + | |||
| + | ==Solution 2== | ||
| + | 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: | ||
| + | |||
| + | <math>n = 1 \implies (3)(2) - (2) = 4</math>, knocking out <math>\textbf{B}</math>, <math>\textbf{C}</math>, and <math>\textbf{E}</math>. | ||
| + | <cmath> </cmath> | ||
| + | <math>n = 2 \implies (4)(3) - (3) = 9</math>, knocking out <math>\textbf{A}</math>. | ||
| + | |||
| + | This leaves <math>\boxed{\textbf{(D)} \text{ a perfect square}}</math> as the only answer choice left. | ||
| + | |||
| + | 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. | ||
| + | |||
| + | ~DBlack2021 | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/6ujfjGLzVoE | ||
| + | |||
| + | ~IceMatrix | ||
| + | |||
| + | ==See Also== | ||
| + | |||
| + | {{AMC12 box|year=2020|ab=B|num-b=5|num-a=7}} | ||
| + | {{MAA Notice}} | ||
Revision as of 08:32, 10 February 2020
Problem 6
For all integers 
the value of
![\[\frac{(n+2)!-(n+1)!}{n!}\]](http://latex.artofproblemsolving.com/e/4/c/e4cf3e216ff29ec12bb85e389993280b1ca40180.png)
is always which of the following?
Solution
We first expand the expression:
![\[\frac{(n+2)!-(n+1)!}{n!} = \frac{(n+2)(n+1)n!-(n+1)n!}{n!}\]](http://latex.artofproblemsolving.com/f/b/b/fbb7fd2ba8842e956cbb39cd25e8f8e1afcb1f1b.png)
We can now divide out a common factor of 
from each term of this expression:
![\[(n+2)(n+1)-(n+1)\]](http://latex.artofproblemsolving.com/a/8/a/a8a064934f7783af35d33f76ddd140390f109856.png)
Factoring out 
, we get ![\[(n+1)(n+2-1) = (n+1)^2\]](http://latex.artofproblemsolving.com/2/9/c/29cf1b77a7720e7818dcafab9161cab373f29495.png)
which proves that the answer is 
.
Solution 2
Factor out an to get: 
Now, without loss of generality, test values of 
until only one answer choice is left valid:
, knocking out 
, 
, and 
.
![\[\]](http://latex.artofproblemsolving.com/5/7/f/57f1406b65f9f2e6a365b7356da731780b42cceb.png)
, knocking out 
.
This leaves as the only answer choice left.
With further testing it becomes clear that for all , 
, proved in Solution 1.
~DBlack2021
Video Solution
~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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
