Difference between revisions of "2022 AMC 10B Problems/Problem 9"
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==Solution== | ==Solution== | ||
| − | Note that <math>\frac{n}{(n+1)!} = \frac{1}{n!} - \frac{1}{(n+1)!}</math>, and therefore this sum is a telescoping sum, which is equivalent to <math>1 - \frac{1}{2022!}</math>. Our answer is <math>1 + 2022 = 2023</math> | + | Note that <math>\frac{n}{(n+1)!} = \frac{1}{n!} - \frac{1}{(n+1)!}</math>, and therefore this sum is a telescoping sum, which is equivalent to <math>1 - \frac{1}{2022!}</math>. Our answer is <math>1 + 2022 = \fbox{D. 2023}</math> |
~mathboy100 | ~mathboy100 | ||
Revision as of 16:10, 17 November 2022
Problem
The sum
![\[\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\dots+\frac{2021}{2022!}\]](http://latex.artofproblemsolving.com/9/e/3/9e3cf91ad68f1674a89bb502c2853024eb12b445.png)
can be expressed as 
, where 
and 
are positive integers. What is 
?
Solution
Note that 
, and therefore this sum is a telescoping sum, which is equivalent to 
. Our answer is 
~mathboy100
See Also
| 2022 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
