Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 5 (Combinatorics))
(→‎Solution 5 (Combinatorics))
Line 46: Line 46:
 
More specifically,  
 
More specifically,  
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
−
\bigcup_{n=2}^{2022}\{\sigma \mid \sigma_{n-1}>\sigma_n\}&=\text{all permutations}\backslash\text{sorted}\\
+
\bigcup_{n=2}^{2022}\{\sigma \mid \sigma_{n-1}>\sigma_n\}&=\{\sigma \mid \sigma_{1}<\sigma_{2}<\ldots<\sigma_{2022}\}^\complement\\
 
\bigsqcup_{n=2}^{2022}\{\sigma \mid \sigma_{n-1}>\sigma_n, \sigma_{n-2}\leq\sigma_{n-1},\ldots, \sigma_{1}\leq\sigma_{2}\}&=\{\sigma \mid \sigma_{1}<\sigma_{2}<\ldots<\sigma_{2022}\}^\complement\\
 
\bigsqcup_{n=2}^{2022}\{\sigma \mid \sigma_{n-1}>\sigma_n, \sigma_{n-2}\leq\sigma_{n-1},\ldots, \sigma_{1}\leq\sigma_{2}\}&=\{\sigma \mid \sigma_{1}<\sigma_{2}<\ldots<\sigma_{2022}\}^\complement\\
 
\#\left(\bigsqcup_{n=2}^{2022}\{\sigma \mid \sigma_{n-1}>\sigma_n, \sigma_{n-2}\leq\sigma_{n-1},\ldots, \sigma_{1}\leq\sigma_{2}\}\right)&=\#\left(\{\sigma \mid \sigma_{1}<\sigma_{2}<\ldots<\sigma_{2022}\}^\complement\right)\\
 
\#\left(\bigsqcup_{n=2}^{2022}\{\sigma \mid \sigma_{n-1}>\sigma_n, \sigma_{n-2}\leq\sigma_{n-1},\ldots, \sigma_{1}\leq\sigma_{2}\}\right)&=\#\left(\{\sigma \mid \sigma_{1}<\sigma_{2}<\ldots<\sigma_{2022}\}^\complement\right)\\

Revision as of 20:25, 24 November 2023

Problem

The sum \[\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{2021}{2022!}\] 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 add $\frac{1}{2022!}$ to the original expression \[\left(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{2021}{2022!}\right)+\frac{1}{2022!}=\left(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\dots+\frac{2020}{2021!}\right)+\frac{1}{2021!}.\] This sum clearly telescopes, thus we end up with $\left(\frac{1}{2!}+\frac{2}{3!}\right)+\frac{1}{3!}=\frac{2}{2!}=1$. Thus the original expression is equal to $1-\frac{1}{2022!}$, and $1+2022=\boxed{\textbf{(D)}\ 2023}$.

~not_slay (+ minor LaTeX edit ~TaeKim)

Solution 3 (Induction)

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

Solution 4

Let $x=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dots+\frac{1}{2022!}.$

Note that \begin{align*} \left(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\dots+\frac{2021}{2022!}\right)+\left(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dots+\frac{1}{2022!}\right)&=\frac{1}{1!}+\frac{2}{2!}+\frac{3}{3!}+\dots+\frac{2022}{2022!}\\ \left(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\dots+\frac{2021}{2022!}\right)+x&=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{2021!}\\ \left(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\dots+\frac{2021}{2022!}\right)+x&=x+1-\frac{1}{2022!}\\ \left(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\dots+\frac{2021}{2022!}\right)&=1-\frac{1}{2022!}. \end{align*} Therefore, the answer is $1+2022=\boxed{\textbf{(D) }2023}.$

~lopkiloinm

Solution 5 (Combinatorics)

Let's examine a list containing all the integers between $1$ and $2022$ inclusive. We want to find the probability of the list being unsorted.

Suppose that we are looking at the first two items in our list. The probability of the first element being greater than the second element is $\frac{1}{2!}$.

When we are looking at the first three items in our list, the probability of the second element being greater than the third element given the first element less than or equal to the second element is $\frac{2}{3!}$.

Similarly, when we are looking at the first four items in our list, the probability of the third element being greater than the fourth element given the second element less than or equal to the third element and the first element less than or equal to the second element is $\frac{3}{4!}$.

More specifically, \begin{align*} \bigcup_{n=2}^{2022}\{\sigma \mid \sigma_{n-1}>\sigma_n\}&=\{\sigma \mid \sigma_{1}<\sigma_{2}<\ldots<\sigma_{2022}\}^\complement\\ \bigsqcup_{n=2}^{2022}\{\sigma \mid \sigma_{n-1}>\sigma_n, \sigma_{n-2}\leq\sigma_{n-1},\ldots, \sigma_{1}\leq\sigma_{2}\}&=\{\sigma \mid \sigma_{1}<\sigma_{2}<\ldots<\sigma_{2022}\}^\complement\\ \#\left(\bigsqcup_{n=2}^{2022}\{\sigma \mid \sigma_{n-1}>\sigma_n, \sigma_{n-2}\leq\sigma_{n-1},\ldots, \sigma_{1}\leq\sigma_{2}\}\right)&=\#\left(\{\sigma \mid \sigma_{1}<\sigma_{2}<\ldots<\sigma_{2022}\}^\complement\right)\\ 2022!\sum_{n=2}^{2022}\frac{n-1}{n!}&=2022!-1\\ \sum_{n=2}^{2022}\frac{n-1}{n!}&=1-\frac{1}{2022!} \end{align*}

This series ends up being the probability of having the list unsorted and that is of course $1-\frac{1}{2022!}$

~lopkiloinm

Solution 6 (Desperate)

Because the fractions get smaller, it is obvious that the answer is less than $1$, so we can safely assume that $a=1$ (this can also be guessed by intuition using similar math problems). Looking at the answer choices, $2018<b<2024$. Because the last term consists of $2022!$ (and the year is $2022$) we can guess that $b=2022$. Adding them yields $1+2022=\boxed{\textbf{(D) }2023}$.

~iluvme and andy_lee

Solution 7 (Partial Fractions)

Knowing that the answer will be in the form $a-\frac{1}{b!}$, we can guess that the sum telescopes. Using partial fractions, we can hope to rewrite $\frac{n-1}{n!}$ as $\frac{A}{(n-1)!}-\frac{B}{n}$. Setting these equal and multiplying by $n!$, we get $n-1=An-B(n-1)!$. Since $An$ is the only term with $n$ with degree $1$, we can conclude that $A=1$. This means that $B=\frac{1}{(n-1)!}$. Substituting, we find that $\frac{n-1}{n!}=\frac{1}{(n-1)!}-\frac{1}{n!}$. This sum clearly telescopes and we obtain $1-\frac{1}{2022!}$. This means that our desired answer is $1+2022=\boxed{\textbf{(D) }2023}.$

~kn07

Solution 8 (Taylor Series)

We calculate the Taylor series error to be $\frac{1}{2023!}$ and this error happens $2023$ times so it is $\frac{2023}{2023!}$ which is $\frac{1}{2022!}$

Video Solution

by Ismail.maths https://www.youtube.com/watch?v=lt34QscjTf4&list=PLmpPPbOoDfgj5BlPtEAGcB7BR_UA5FgFj

https://youtu.be/4vdVGYXGvzg

- Whiz

Video Solution by Interstigation

https://youtu.be/_KNR0JV5rdI?t=1102

Video Solution by paixiao

https://www.youtube.com/watch?v=MRwZ4F7Scog

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_10B_Problems/Problem_9&oldid=205621"