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

Revision as of 18:18, 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

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 $\tfrac{1}{2!} + \tfrac{2}{3!} + \cdots \tfrac{n-1}{n! = 1 - \tfrac{1}{n!}$ (Error compiling LaTeX. Unknown error_msg) 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.

@@ Line 17: / Line 17: @@
 ~not_slay
+==Solution 3 (Induction)==
+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>.
+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 ==
 {{AMC10 box|year=2022|ab=B|num-b=8|num-a=10}}
 {{MAA Notice}}

2022 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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