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Difference between revisions of "2022 AMC 10B Problems/Problem 9"

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~not_slay
 
~not_slay
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==Solution 3 (Induction)==
 
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By looking for a pattern, we see that <math>\tfrac{1}{2!} = 1 - \tfrac{1}{2!}</math> and <math>\tfrac{1}{2!} + \tfrac{2}{3!} = \tfrac{5}{6} = 1 - \tfrac{1}{3!}</math>, so we can conclude by engineer's induction that the sum in the problem is equal to <math>1 - \tfrac{1}{2022!}</math>, for an answer of <math>\boxed{\textbf{(D)}\ 2023}</math>.
 
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This can be proven with actual induction as well; we have already established <math>2</math> base cases, so now assume that <math>\tfrac{1}{2!} + \tfrac{2}{3!} + \cdots \tfrac{n-1}{n! = 1 - \tfrac{1}{n!}</math> for <math>n = k</math>. For <math>n = k + 1</math> we get <math>\tfrac{1}{2!} + \tfrac{2}{3!} + \cdots \tfrac{n-1}{n!} + \tfrac{n}{(n+1)!} = 1 - \tfrac{1}{n!} + \tfrac{n}{(n+1)!} = 1 - \tfrac{n+1}{(n+1)!} + \tfrac{n}{(n+1)!} = 1 - \tfrac{1}{(n+1)!}</math>, completing the proof.
 
  
 
== 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 19:18, 18 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 1

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

Solution 2

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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