Problem:
The sum
2!1+3!2+4!3+⋯+2022!2021
can be expressed as a−b!1, where a and b are positive integers. What is a+b?
Answer Choices:
A. 2020
B. 2021
C. 2022
D. 2023
E. 2024
Solution:
Notice that for n≥2,
n!n−1=(n−1)!1−n!1.
Therefore the given sum telescopes:
1!1−2!1+2!1−3!1+3!1−4!1+⋯+2021!1−2022!1=1−2022!1
Thus a=1,b=2022, and a+b=(D)2023.
The problems on this page are the property of the MAA's American Mathematics Competitions