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

Revision as of 03:02, 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 $x=\left(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dots+\frac{1}{2022!}\right)$ $\begin{align*} \left(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\dots+\frac{2021}{2022!}\right)+\left(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dots+\frac{1}{2022!}\right)&=\left(\frac{1}{1!}+\frac{2}{2!}+\frac{3}{3!}+\dots+\frac{2022}{2022!}\right)\\ \left(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\dots+\frac{2021}{2022!}\right)+x&=\left(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{2021!}\right)\\ \left(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\dots+\frac{2021}{2022!}\right)+x&=x+1-\frac{1}{2022!}\\ \left(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\dots+\frac{2021}{2022!}\right)&=1-\frac{1}{2022!} \end{align*}$ ~lopkiloinm

Solution 5(Combinatorics)

Suppose you are picking a permutation of $2022$ elements. Suppose that the correct order of the permutation is $x_1,x_2,x_3,\ldots,x_2021,x_2022$ It ends up being the probability of making every permutation except one permutation so it is $1-\frac{1}{2022!}$ . ~lopkiloinm

Video Solution

https://youtu.be/4vdVGYXGvzg

- Whiz

@@ Line 30: / Line 30: @@
 \end{align*}</cmath>~lopkiloinm
 ==Solution 5(Combinatorics)==
-Suppose you are picking a permutation of <math>2022</math> elements. It ends up being the probability of making every permutation except one permutation so it is <math>1-\frac{1}{2022!}</math>. ~lopkiloinm
+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>
+It ends up being the probability of making every permutation except one permutation so it is <math>1-\frac{1}{2022!}</math>. ~lopkiloinm
 == Video Solution ==

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