2022 AMC 10B Problem 9

Question

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

Look for a difference involving consecutive factorial reciprocals

Rewrite each term so the sum telescopes

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 .

Problem 9 in other years