HMMT 二月 2003 · 冲刺赛 · 第 30 题

HMMT February 2003 — Guts Round — Problem 30

[10] The sequence a , a , a , . . . of real numbers satisfies the recurrence 1 2 3 2 a − a + 2 a n − 1 n n a = . n +1 a + 1 n − 1 Given that a = 1 and a = 7, find a . 1 9 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HARVARD-MIT MATHEMATICS TOURNAMENT, MARCH 15, 2003 — GUTS ROUND

解析

The sequence a , a , a , . . . of real numbers satisfies the recurrence 1 2 3 2 a − a + 2 a n − 1 n n a = . n +1 a + 1 n − 1 Given that a = 1 and a = 7, find a . 1 9 5 Solution: 3 2 2 Let b = a +1. Then the recurrence becomes b − 1 = ( b − b ) /b = b /b − 1, n n n +1 n − 1 n − 1 n − 1 n n 2 so b = b /b . It follows that the sequence ( b ) is a geometric progression, from n +1 n − 1 n n 2 which b = b b = 2 · 8 = 16 ⇒ b = ± 4. However, since all b are real, they either 1 9 5 n 5 alternate in sign or all have the same sign (depending on the sign of the progression’s common ratio); either way, b has the same sign as b , so b = 4 ⇒ a = 3. 5 1 5 5