HMMT 二月 2014 · 冲刺赛 · 第 31 题
HMMT February 2014 — Guts Round — Problem 31
题目详情
- [ 20 ] Compute ✓ ✓ ◆◆ 1007 2014 X ⇡ k cos . 1007 k =1 2 2 10 b 10 a
解析
- [ 20 ] Compute ( ( )) 1007 2014 ∑ πk cos . 1007 k =1 2013 2014(1+ ) 2 πi ( ) 1007 2014 πk 1 k − k 2014 Answer: Let ω = e . We have ω = 1. Note that cos( ) = ( ω + ω ). 2014 2 1007 2 Our desired expression is 1007 ∑ 1 k − k 2014 ( ω + ω ) 2014 2 k =1 Using binomial expansion and switching the order of the resulting summation, this is equal to ( ) 2014 1007 ∑ ∑ 1 2014 2014 − 2 j k ( ω ) 2014 2 j j =0 k =1 . 2014 − 2 j Note that unless ω = 1, the summand 1007 ∑ 2014 − 2 j k ( ω ) k =1 is the sum of roots of unity spaced evenly around the unit circle in the complex plane (in particular the 1007th, 19th, and 53rd roots of unity), so it is zero. Thus, we must only sum over those j for which 2014 − 2 j ω = 1, which holds for j = 0 , 1007 , 2014. This yields the answer ( ( )) ( ( ) ) 2013 2014 1 + 1 2014 1007 1007 + 1007 + 1007 = . 2014 2014 2 1007 2 10 b 10 a 2 2