AMC 10B 2022 (Problem 9)The sum 12!+23!+34!+⋯+20212022!\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \cdots + \frac{2021}{2022!}2!1+3!2+4!3+⋯+2022!2021 can be expressed as a−1b!a - \frac{1}{b!}a−b!1, where aaa and bbb are positive integers. What is a+ba+ba+b?(A) 2020\text{(A)}\;2020(A)2020(B) 2021\text{(B)}\;2021(B)2021(C) 2022\text{(C)}\;2022(C)2022(D) 2023\text{(D)}\;2023(D)2023(E) 2024\text{(E)}\;2024(E)2024Related TopicsToolkit 5 — Telescoping seriesHints (3)Hint 1Use 5. Telescoping Series.Hint 2k(k+1)!=1k!−1(k+1)!\frac{k}{(k+1)!} = \frac{1}{k!} - \frac{1}{(k+1)!}(k+1)!k=k!1−(k+1)!1Hint 312!+23!+⋯+20212022!=(11!−12!)+(12!−13!)+⋯+(12021!−12022!)=1−12022!\frac{1}{2!}+\frac{2}{3!}+\cdots+\frac{2021}{2022!} = \left(\frac{1}{1!}-\frac{1}{2!}\right)+\left(\frac{1}{2!}-\frac{1}{3!}\right)+\cdots+\left(\frac{1}{2021!}-\frac{1}{2022!}\right)=1-\frac{1}{2022!}2!1+3!2+⋯+2022!2021=(1!1−2!1)+(2!1−3!1)+⋯+(2021!1−2022!1)=1−2022!1. Therefore a=1a=1a=1 and b=2022b=2022b=2022, so a+b=2023a+b=2023a+b=2023.Final Answer(D) 2023