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Difference between revisions of "2022 AMC 10B Problems/Problem 9"
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~mathboy100
 
~mathboy100
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We have <math>(\frac{1}{2!}+\frac{2}{3!}+\dots+\frac{2021}{2022!})+\frac{1}{2022!}=(\frac{1}{2!}+\frac{2}{3!}+\dots+\frac{2020}{2021!})+\frac{1}{2021!}</math> from canceling a 2022 from <math>\frac{2021+1}{2022!}</math>.
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This sum clearly telescopes, thus we end up with <math>(\frac{1}{2!}+\frac{2}{3!})+\frac{1}{3!}=\frac{2}{2!}=1</math>. Thus the original equation is equal to <math>1-\frac{1}{2022}</math>, and <math>1+2022=2023</math>. <math>\boxed{\textbf{(D)}\ 2023}</math>.
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~not_slay
  
 
== 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 20:10, 17 November 2022

Problem

The sum \[\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\dots+\frac{2021}{2022!}\]can be expressed as $a-\frac{1}{b!}$, where $a$ and $b$ are positive integers. What is $a+b$?

$\textbf{(A)}\ 2020 \qquad\textbf{(B)}\ 2021 \qquad\textbf{(C)}\ 2022 \qquad\textbf{(D)}\ 2023 \qquad\textbf{(E)}\ 2024$

Solution

Note that $\frac{n}{(n+1)!} = \frac{1}{n!} - \frac{1}{(n+1)!}$, and therefore this sum is a telescoping sum, which is equivalent to $1 - \frac{1}{2022!}$. Our answer is $1 + 2022 = \boxed{\textbf{(D)}\ 2023}$.

~mathboy100


We have $(\frac{1}{2!}+\frac{2}{3!}+\dots+\frac{2021}{2022!})+\frac{1}{2022!}=(\frac{1}{2!}+\frac{2}{3!}+\dots+\frac{2020}{2021!})+\frac{1}{2021!}$ from canceling a 2022 from $\frac{2021+1}{2022!}$. This sum clearly telescopes, thus we end up with $(\frac{1}{2!}+\frac{2}{3!})+\frac{1}{3!}=\frac{2}{2!}=1$. Thus the original equation is equal to $1-\frac{1}{2022}$, and $1+2022=2023$. $\boxed{\textbf{(D)}\ 2023}$.

~not_slay

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

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