Difference between revisions of "2022 AMC 10B Problems/Problem 9"
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<math> \textbf{(A)}\ 2020 \qquad\textbf{(B)}\ 2021 \qquad\textbf{(C)}\ 2022 \qquad\textbf{(D)}\ 2023 \qquad\textbf{(E)}\ 2024</math> | <math> \textbf{(A)}\ 2020 \qquad\textbf{(B)}\ 2021 \qquad\textbf{(C)}\ 2022 \qquad\textbf{(D)}\ 2023 \qquad\textbf{(E)}\ 2024</math> | ||
| − | ==Solution= | + | ==Solution== |
== See Also == | == See Also == | ||
{{AMC10 box|year=2022|ab=B|num-b=8|num-a=10}} | {{AMC10 box|year=2022|ab=B|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 14:10, 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
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. 
