PUMaC 2015 · 代数（A 组） · 第 4 题
PUMaC 2015 — Algebra (Division A) — Problem 4
题目详情
- [ 4 ] Define the sequence a as follows: a = 1 , a = 2015, and a = for n > 2. i 1 2 n a + na n − 1 n − 2 What is the least k such that a < a ? k k − 1
解析
- [ 4 ] Define the sequence a as follows: a = 1 , a = 2015, and a = for n > 2. i 1 2 n a + na n − 1 n − 2 What is the least k such that a < a ? k k − 1 Solution: ∑ a a n a n − 1 n − 2 1 1 1 k − 1 The recursion is equivalent to = + = + . The first k for which > 1 i =3 a a n 2015 i a n n − 1 k occurs when k = 7 by a simple computation of the sum. Author: Bill Huang