HMMT 二月 2011 · 冲刺赛 · 第 12 题

HMMT February 2011 — Guts Round — Problem 12

[ 8 ] A sequence of integers { a } is defined as follows: a = i for all 1 ≤ i ≤ 5, and a = a a · · · a − 1 i i i 1 2 i − 1 2011 ∑ 2 for all i > 5. Evaluate a a · · · a − a . 1 2 2011 i i =1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . th 14 HARVARD-MIT MATHEMATICS TOURNAMENT, 12 FEBRUARY 2011 — GUTS ROUND b + c − a

解析

[ 8 ] A sequence of integers { a } is defined as follows: a = i for all 1 ≤ i ≤ 5, and a = a a · · · a − 1 i i i 1 2 i − 1 2011 ∑ 2 for all i > 5. Evaluate a a · · · a − a . 1 2 2011 i i =1 Answer: − 1941 For all i ≥ 6, we have a = a a · · · a − 1. So i 1 2 i − 1 a = a a · · · a − 1 i +1 1 2 i = ( a a · · · a ) a − 1 1 2 i − 1 i = ( a + 1) a − 1 i i 2 = a + a − 1 . i i 2 Therefore, for all i ≥ 6, we have a = a − a + 1, and we obtain that i +1 i i 2011 ∑ 2 a a · · · a − a 1 2 2011 i i =1 5 2011 ∑ ∑ 2 2 = a + 1 − a − a 2012 i i i =1 i =6 5 2011 ∑ ∑ 2 = a + 1 − i − ( a − a + 1) 2012 i +1 i i =1 i =6 = a + 1 − 55 − ( a − a + 2006) 2012 2012 6 = a − 2060 6 = − 1941 b + c − a