Difference between revisions of "2020 AMC 12B Problems/Problem 6"
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==See Also== | ==See Also== | ||
Latest revision as of 14:02, 8 June 2023
Contents
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/8/6/2/862026a8218274b6eb950d238448ee739d863884.png)
We can now divide out a common factor of 
from each term of the numerator:
![\[(n+2)(n+1)-(n+1).\]](http://latex.artofproblemsolving.com/a/6/5/a65f6dcdd7e879078eb8acdcb6c41ffcc21f8f0e.png)
Factoring out 
we get  = (n+1)^2,\]](http://latex.artofproblemsolving.com/7/0/c/70c754fc426f757786accf6318b93d7bc4d1d175.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 
and 
knocking out 
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. The condition 
was added most likely to encourage picking 
and discourage substituting smaller values into 
~DBlack2021 (Solution)
~MRENTHUSIASM (Edits in Logic)
~Countmath1 (Minor Edits in Formatting)
Video Solution (HOW TO THINK CRITICALLY!!!)
~Education, the Study of Everything
Video Solution
https://youtu.be/ba6w1OhXqOQ?t=2234
Video Solution
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. 
