Difference between revisions of "2020 AMC 12B Problems/Problem 6"
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==Solution 2== | ==Solution 2== | ||
| − | + | In the numerator, we factor out an <math>n!</math> to get <cmath>\frac{(n+2)!-(n+1)!}{n!} = (n+2)(n+1)-(n+1).</cmath> 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)},\textbf{(C)},</math> and <math>\textbf{(E)}</math>. | <math>n = 1 \implies (3)(2) - (2) = 4</math>, knocking out <math>\textbf{(B)},\textbf{(C)},</math> and <math>\textbf{(E)}</math>. | ||
Revision as of 01:46, 9 June 2021
Problem
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 1
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
In the numerator, we factor out an to get ![\[\frac{(n+2)!-(n+1)!}{n!} = (n+2)(n+1)-(n+1).\]](http://latex.artofproblemsolving.com/6/6/0/6605f384fc7eb446d310e1fc0b187538f98f32c1.png)
Now, without loss of generality, test values of 
until only one answer choice is left valid:
, knocking out 
and 
.
, knocking out 
This leaves as the only answer choice left.
This solution does not consider the condition 
The reason is that, with further testing it becomes clear that for all 
we get ![\[(n+2)(n+1)-(n+1) = (n+1)^{2},\]](http://latex.artofproblemsolving.com/2/2/7/227c9dcc1a25f21fe49c32d711cf9a9fb6086260.png)
as proved in Solution 1. We have now revealed that the condition 
is insignificant.
~DBlack2021 (Solution Writing)
~MRENTHUSIASM (Edits in Logic)
Video Solution
https://youtu.be/ba6w1OhXqOQ?t=2234
~ pi_is_3.14
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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
