2022 AMC 10 B

Complete problem set with solutions and individual problem pages

Problem 9 Easy

The sum \frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{2021}{2022!} can be expressed as a-\dfrac{1}{b!}  ,  where a and b are positive integers. What is a+b?

  • A.

    2020

  • B.

    2021

  • C.

    2022

  • D.

    2023

  • E.

    2024

Answer:D

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=2023.

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