Difference between revisions of "2020 AMC 12B Problems/Problem 6"
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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>\(n+1)(n+2-1) = (n+1)^2</cmath> | This expression can be shown as <cmath>\(n+1)(n+2-1) = (n+1)^2</cmath> | ||
| − | which proves that the answer is | + | which proves that the answer is <math>\textbf{(D)}</math>. |
| + | |||
| + | ==Problem 6== | ||
| + | |||
| + | For all integers <math>n \geq 9,</math> the value of | ||
| + | <cmath>\frac{(n+2)!-(n+1)!}{n!}</cmath>is always which of the following? | ||
| + | |||
| + | <math>\textbf{(A) } \text{a multiple of }4 \qquad \textbf{(B) } \text{a multiple of }10 \qquad \textbf{(C) } \text{a prime number} \\ \textbf{(D) } \text{a perfect square} \qquad \textbf{(E) } \text{a perfect cube}</math> | ||
Revision as of 19:22, 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\] (Error compiling LaTeX. Unknown error_msg)
which proves that the answer is 
.
Problem 6
For all integers the value of
is always which of the following?
