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 TopicsCoreToolkit 5 — Telescoping seriesHints (3)Hint 1k(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 212!+23!+34!+⋯+20212022!\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{2021}{2022!}2!1+3!2+4!3+⋯+2022!2021=11!−12!=\quad\frac{1}{1!}-\frac{1}{2!}=1!1−2!1+12!−13!\qquad+\frac{1}{2!}-\frac{1}{3!}+2!1−3!1+13!−14!\qquad+\frac{1}{3!}-\frac{1}{4!}+3!1−4!1⋮\qquad\vdots⋮+12021!−12022!\qquad+\frac{1}{2021!}-\frac{1}{2022!}+2021!1−2022!1=1−12022!=1-\frac{1}{2022!}=1−2022!1Hint 3a−1b!=1−12022!a-\frac{1}{b!}=1-\frac{1}{2022!}a−b!1=1−2022!1⟹a=1,b=2022\Longrightarrow a=1,\qquad b=2022⟹a=1,b=2022⟹a+b=2023\Longrightarrow a+b=2023⟹a+b=2023Final Answer(D) 202320232023Related Problems (4)AMC 8 2022 (Problem 8)AMC 12B 2025 (Problem 7)MATHCOUNTS 2026 State Sprint Round (Problem 24)AIME II 2025 (Problem 4)