PUMaC 2016 · 团队赛 · 第 4 题
PUMaC 2016 — Team Round — Problem 4
题目详情
- (4) For x > 1, let f ( x ) = log ( x + log ( x + log ( x + . . . ))). Compute 2 2 2 10 ∑ − 1 f ( k ) . k =2
解析
- If f ( x ) = k , we have 2 − x = k , so x = 2 − k . Thus, we have 10 10 10 10 ∑ ∑ ∑ ∑ − 1 k k 11 f ( k ) = (2 − k ) = 2 − k = 2 − 4 − 54 = 2048 − 58 = 1990 , k =2 k =2 k =2 k =2 as desired. Problem written by Eric Neyman.