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PUMaC 2021 · 代数(B 组) · 第 7 题

PUMaC 2021 — Algebra (Division B) — Problem 7

专题
Discrete Math / 离散数学
难度
L3
来源
PUMaC

题目详情

  1. Consider the sum 2021 X 2 πj S = | sin | . 2021 j =1 cπ The value of S can be written as tan( ) for some relatively prime positive integers c, d, d satisfying 2 c < d . Find the value of c + d .
解析
  1. Consider the sum 2021 X 2 πj S = | sin | . 2021 j =1 cπ The value of S can be written as tan( ) for some relatively prime positive integers c, d, d satisfying 2 c < d . Find the value of c + d . Proposed by: Frank Lu Answer: 3031 We know that the terms are positive for j = 1 , 2 , . . . , 1010 , and that they are negative for j = 2 π (2021 − j ) 2 πj 1011 , 1012 , . . . , 2020 , and 0 for j = 2021 . Furthermore, notice that sin = − sin , 2021 2021 1010 P 2 πj so therefore this sum can be written as 2 sin . 2021 j =1 2 1010 P 2 iπj 2021 From here, we can think about this as the imaginary part of the sum of exponentials 2 e . j =1 2022 iπ 2 iπ 2021 2021 e − e From the formula for a geometric series, we may write this as 2 . But recall that 2 iπ 2021 e − 1 iπ 2 iπ 2021 2021 iπ − e − e e = − 1 , meaning that this is just equal to 2 . From here, we may further simplify 2 iπ 2021 e − 1 iπ 2021 − e this as 2 . iπ 2021 e − 1 Finally, observe that if we add 1 to this, we will not change the imaginary part of this value. iπ 2021 − e +1 π π π 1010 π But adding one yields us with the fraction = cot( ) = tan( − ) = tan( ) . iπ 4042 2 4042 2021 2021 e − 1 Our answer is thus 1010 + 2021 = 3031 .