2020 AMC 12B Problems/Problem 6: Difference between revisions
| Line 7: | Line 7: | ||
==Solution== | ==Solution== | ||
<cmath>\frac{(n+2)!-(n+1)!}{n!} | We first expand the expression: | ||
<cmath>\frac{(n+2)!-(n+1)!}{n!} = \frac{(n+2)(n+1)n!-(n+1)n!}{n!}</cmath> | |||
which proves that the answer is <math>\textbf{(D)}</math>. | 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) a perfect square}}</math>. | |||
Revision as of 19:35, 7 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 
.
