Difference between revisions of "Random Problem"

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Problem
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==Problem==
 
The sum<cmath>\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{2021}{2022!}</cmath>can be expressed as <math>a-\frac{1}{b!}</math>, where <math>a</math> and <math>b</math> are positive integers. What is <math>a+b</math>?
 
The sum<cmath>\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{2021}{2022!}</cmath>can be expressed as <math>a-\frac{1}{b!}</math>, where <math>a</math> and <math>b</math> are positive integers. What is <math>a+b</math>?
  
 
<math>\textbf{(A)}\ 2020 \qquad\textbf{(B)}\ 2021 \qquad\textbf{(C)}\ 2022 \qquad\textbf{(D)}\ 2023 \qquad\textbf{(E)}\ 2024</math>
 
<math>\textbf{(A)}\ 2020 \qquad\textbf{(B)}\ 2021 \qquad\textbf{(C)}\ 2022 \qquad\textbf{(D)}\ 2023 \qquad\textbf{(E)}\ 2024</math>
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==Solution==
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???

Revision as of 18:15, 27 January 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

???