Difference between revisions of "2020 AMC 12B Problems/Problem 6"
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<cmath>\frac{(n+2)!-(n+1)!}{n!}</cmath> can be simplified by common denominator n!. | <cmath>\frac{(n+2)!-(n+1)!}{n!}</cmath> can be simplified by common denominator n!. | ||
Therefore, <cmath>\frac{(n+2)!-(n+1)!}{n!} = (n+2)(n+1)-(n+1)</cmath> | Therefore, <cmath>\frac{(n+2)!-(n+1)!}{n!} = (n+2)(n+1)-(n+1)</cmath> | ||
| − | This expression can be shown as <cmath> | + | This expression can be shown as <cmath>(n+1)(n+2-1) = (n+1)^2</cmath> |
which proves that the answer is <math>\textbf{(D)}</math>. | which proves that the answer is <math>\textbf{(D)}</math>. | ||
Revision as of 20:34, 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
can be simplified by common denominator n!.
Therefore, ![\[\frac{(n+2)!-(n+1)!}{n!} = (n+2)(n+1)-(n+1)\]](http://latex.artofproblemsolving.com/c/2/2/c2267385bac56e31aba19576629bceb9f07127e6.png)
This expression can be shown as ![\[(n+1)(n+2-1) = (n+1)^2\]](http://latex.artofproblemsolving.com/2/9/c/29cf1b77a7720e7818dcafab9161cab373f29495.png)
which proves that the answer is 
.
