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

2022 AMC 10B Problem 9 — Solution

2022 AMC 10B Problem 9

Attempt Problem 9 of the 2022 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2022 AMC 10B solutions, or check the answer key.

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

$\ 2020$

$\ 2021$

$\ 2022$

$\ 2023$

$\ 2024$

Answer: D

Concepts:telescoping factorial induction

Difficulty rating: 1220

Solution:

Each term telescopes because $\frac{k}{(k+1)!}=\frac{1}{k!}-\frac{1}{(k+1)!}.$ Summing makes every intermediate factorial reciprocal cancel, leaving $1-\frac1{2022!}.$ Hence $a=1, b=2022,$ and $a+b=2023.$

Thus, our answer is D .

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