Skip to main content

2022 AMC 10B Problem 9

Problem 9 of 25EasierAlgebraCounting & Probability

The sum 12!+23!+34!+⋯+20212022!\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+\cdots+\dfrac{2021}{2022!} can be expressed as a−1b!,a-\dfrac{1}{b!}, where aa and bb are positive integers. What is a+b?a+b?

Answer choices

Show solution

Solution

Each term telescopes because k(k+1)!=1k!−1(k+1)!.\frac{k}{(k+1)!}=\frac{1}{k!}-\frac{1}{(k+1)!}. Summing makes every intermediate factorial reciprocal cancel, leaving 1−12022!.1-\frac1{2022!}. Hence a=1,b=2022,a=1, b=2022, and a+b=2023.a+b=2023. Thus, our answer is D .

More practice

Concepts: telescoping · factorial · induction

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.