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

Revision as of 04:51, 22 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 $\left(\frac{1}{2!}+\frac{2}{3!}+\dots+\frac{2021}{2022!}\right)+\frac{1}{2022!}=\left(\frac{1}{2!}+\frac{2}{3!}+\dots+\frac{2020}{2021!}\right)+\frac{1}{2021!}$ from canceling a 2022 from $\frac{2021+1}{2022!}$ . 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 equation is equal to $1-\frac{1}{2022}$ , and $1+2022=2023$ . $\boxed{\textbf{(D)}\ 2023}$ .

~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)

Suppose you are picking a permutation of $2022$ elements. Suppose that the correct order of the permutation is $x_1,x_2,x_3,\ldots,x_{2021},x_{2022}$ We want to find the probability of picking the permutation in the wrong order.

Suppose that we have picked everything to the correct order except our last $2$ elements. That is we have $x_1,x_2,x_3,\ldots,x_{2019},x_{2020}$ We want pick the next element such that it does not equal to $x_{2021}$ . There are $1$ ways to choose that, so we add $\frac{1}{2!}$ to the probability.

Suppose that we have picked everything to the correct order except our last $3$ elements. That is we have $x_1,x_2,x_3,\ldots,x_{2018},x_{2019}$ We want pick the next element such that it does not equal to $x_{2020}$ . There are $2$ ways to choose that, so we add $\frac{2}{3!}$ to the probability.

This series ends up being the probability of making a permutation in the wrong order and that is of course $1-\frac{1}{2022!}$ ~lopkiloinm

Video Solution

https://youtu.be/4vdVGYXGvzg

- Whiz

@@ Line 35: / Line 35: @@
 ~lopkiloinm
-==Solution 5(Combinatorics)==
+==Solution 5 (Combinatorics)==
 Suppose you are picking a permutation of <math>2022</math> elements. Suppose that the correct order of the permutation is
 <cmath>x_1,x_2,x_3,\ldots,x_{2021},x_{2022}</cmath>

2022 AMC 10B (Problems • Answer Key • Resources)
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