HMMT 二月 2014 · 冲刺赛 · 第 31 题

英文解析

31. [ 20 ] Compute
    ( ( ))
    1007
    2014
    ∑
    cos .πk
    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
    ∑
    k − k 20141
    ( ω + ω )
    2014
    k =12
    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 jj =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 particularthe 1007 th, 19 th, and 53 rd roots of unity), so it is zero. Thus, we must only sum over those j for which
    2014 − 2 jω = 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