HMMT 十一月 2018 · 冲刺赛 · 第 25 题

HMMT November 2018 — Guts Round — Problem 25

[ 13 ] Let a , a , . . . and b , b , . . . be geometric sequences with common ratios r and r , respectively, 0 1 0 1 a b such that ( ) ( ) ∞ ∞ ∞ ∞ ∞ ∑ ∑ ∑ ∑ ∑ 2 2 a = b = 1 and a b = a b . i i i i i i i =0 i =0 i =0 i =0 i =0 Find the smallest real number c such that a < c must be true. 0

解析

[ 13 ] Let a , a , . . . and b , b , . . . be geometric sequences with common ratios r and r , respectively, 0 1 0 1 a b such that ! ! 1 1 1 1 1 X X X X X 2 2 a = b = 1 and a b = a b . i i i i i i i =0 i =0 i =0 i =0 i =0 Find the smallest real number c such that a < c must be true. 0 Proposed by: Handong Wang 4 Answer: 3 P 1 a 0 Let a = a and b = b . From a = = 1 we have a = 1 r and similarly b = 1 r . This 0 0 i 0 a 0 b i =0 1 r a 2 P P P P 2 2 1 a 1 1 1 a a a 2 2 2 0 means a = = = = , so a b = a b yields 2 i i i i i i =0 i =0 i =0 i =0 1 r (1 r )(1+ r ) a (2 a ) 2 a a a a a b ab · = . 2 a 2 b 1 (1 a )(1 b ) Since the numerators are equal, the denominators must be equal, which when expanded gives 2 ab 3 a 3 b + 4 = 0, which is equivalent to (2 a 3)(2 b 3) = 1. But note that 0 < a, b < 2 since we need the sequences to converge (or | r | , | r | < 1), so then 3 < 2 b 3 < 1, and thus 2 a 3 > 1 (impossible) a b 1 4 or 2 a 3 < . Hence a < , with equality when b approaches 0. 3 3