PUMaC 2017 · 代数（A 组） · 第 5 题
PUMaC 2017 — Algebra (Division A) — Problem 5
题目详情
- Let f ( x ) = x , and for each n ≥ 0, let f ( x ) = f ( x (3 − 2 x )). Find the smallest real number 0 n +1 n that is at least as large as 2017 2017 ∑ ∑ f ( a ) + f (1 − a ) n n n =0 n =0 for all a ∈ [0 , 1].
解析
- If a + b = 1, then a + b = a − ab + b and a + 2 ab + b = 1, so 2 2 2 2 2 3 3 2 2 2 a (3 − 2 a ) + b (3 − 2 b ) = a + 2( a + b − ( a + b )) + b = a + 2 ab + b = 1 . Therefore, by induction, f ( a ) + f ( b ) = 1 for all n , so the sum is always 2018 . n n Problem written by Matt Tyler