PUMaC 2017 · 代数（A 组） · 第 8 题
PUMaC 2017 — Algebra (Division A) — Problem 8
题目详情
- Let 2017 p = q 1 2 − 2 3 − 1 2 − 1 2 − 2 3 − 1 2 − 1 2 − 1 2 − 2 3 − 1 2 − 2 − · · · where p and q are relatively prime positive integers. Find p + q . 1
解析
- Let a , a , a , . . . be a sequence of real numbers, and define the sequence b , b , b , . . . of real 1 2 3 0 1 2 numbers inductively by b = 1 and b = a b for all n > 0. We will show by induction on 0 n n n − 1 N that 〈 〉 N ∑ 1 − a − a 1 N b = , . . . . , , n 1 1 + a 1 + a 1 N n =0 where 〈·〉 denotes a continued fraction, e.g. 〈 〉 1 − 3 5 1 , , = . − 3 2 4 6 2 + 5 4+ 6 2 If N = 0, then the result is clear, so we may assume N > 0. Define the sequence c , c , c , . . . 0 1 2 inductively by c = 1 and c = a c for all n > 0. By the inductive hypothesis, we may 0 n n +1 n − 1 assume that 〈 〉 N − 1 ∑ 1 − a − a 2 N c = , . . . . , . n 1 1 + a 1 + a 2 N n =0 Let 〈 〉 − a − a 2 N x = . . . . , , 1 + a 1 + a 2 N so that N N − 1 ∑ ∑ b = 1 + a c n 1 n n =0 n =0 a 1 = 1 + 1 + x 1 = a 1 1 − 1+ a + x 1 〈 〉 1 − a − a 1 N = , . . . . , , 1 1 + a 1 + a 1 N as claimed. ∑ Therefore, when b converges, the infinite continued fraction (taking N → ∞ ) is well- n n ≥ 0 defined. Now, we can write 〈 〉 1 1 − 1 − 1 − 2 − 1 − 1 − 2 − 1 − 1 − 1 − 2 − 1 = , , , , , , , , , , , , . . . p 1 − 1 2 3 2 2 3 2 2 2 3 2 2 2017 q 〈 〉 1 − 1 − 1 / 2 − 1 − 1 − 1 / 2 − 1 − 1 − 1 − 1 / 2 − 1 − 1 = , , , , , , , , , , , , . . . , 1 2 3 / 2 2 2 3 / 2 2 2 2 3 / 2 2 2 so the corresponding sequence is { a } = [1 , 1 / 2 , 1 , 1 , 1 / 2 , 1 , 1 , 1 , 1 / 2 , 1 , 1 , . . . ] and n n> 0 N ∑ 2 3 4 5 6 b = + + + + + . . . = 6 . n 1 2 4 8 16 n =0 We then find p 5 = −→ p + q = 10091 . 2017 q 6 Problem written by Bill Huang If you believe that any of these answers is incorrect, or that a problem had multiple reasonable interpretations or was incorrectly stated, you may appeal at http://tinyurl.com/PUMaCappeal2017. All appeals must be in by 1 PM to be considered. 3