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HMMT 二月 2010 · TEAM2 赛 · 第 7 题

HMMT February 2010 — TEAM2 Round — Problem 7

专题
Discrete Math / 离散数学
难度
L3
来源
HMMT

题目详情

  1. [ 30 ] Evaluate 2010 ∑ 2 cos ( k ) k =1
解析
  1. [ 30 ] Evaluate 2010 ∑ 2 cos ( k ) k =1 sin(4021) − sin(1) 1+cos(2 x ) 2 Answer: 1005 + We use the identity cos ( k ) = . Then our expression 4 sin(1) 2 (cos(2)+ ... +cos(4020)) evalutes to 1005 + . 2 To evaluate cos(2) + · · · + cos(4020), let y = cos(2) + · · · + cos(4020) ⇒ y (sin(1)) = cos(2) sin(1) + sin( x +1) − sin( x − 1) . . . + cos(4020) sin(1). Observe that for any x , cos( x ) sin(1) = . Then y (sin(1)) = 2 sin(3) − sin(1) sin(5) − sin(3) sin 4021 − sin(4019)
    • . . . + . This is clearly a telescoping sum; we get y (sin(1)) = 2 2 2 sin(4021) − sin(1) sin(4021) − sin(1) . Then we have the desired y = . Then our original expression evaluates 2 2 sin(1) sin(4021) − sin(1) to 1005 + . 4 sin(1)