PUMaC 2023 · 代数（B 组） · 第 2 题

PUMaC 2023 — Algebra (Division B) — Problem 2

The sum 2023 X 2 m 4 2 m + m + 1 m =1 a can be expressed as for relatively prime positive integers a, b . Find the remainder when a + b b is divided by 1000. 2 2 2 2 2 2

解析

The sum 2023 X 2 m 4 2 m + m + 1 m =1 a can be expressed as for relatively prime positive integers a, b . Find the remainder when a + b b is divided by 1000. Proposed by Sunay Joshi Answer: 105 2 N − N If the sum runs from m = 1 to N − 1, then it has the closed form , where the numerator 2 N − N +1 4 2 2 and denominator are relatively prime. This is by telescoping: note m + m + 1 = ( m − m + 2 2 m 1 1 1)( m + m + 1), so partial fraction decomposition gives = − . 4 2 m + m +1 m ( m − 1)+1 m ( m +1)+1 2 1 1 N − N Accordingly, the sum telescopes into − = , as claimed. Because 2 1 · 0+1 ( N − 1) · N +1 N − N +1 2 2 2 N − N and N − N + 1 differ by 1, they’re relatively prime, so a + b = 2( N − N ) + 1. Setting N = 2024, we find the answer of 105. 2 2 2 2 2 2