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2022 AMC 10B Problems/Problem 9

Revision as of 12:38, 19 November 2022 by Taekim (talk | contribs) (→‎Solution 2)

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

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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