HMMT 二月 2003 · 冲刺赛 · 第 30 题
HMMT February 2003 — Guts Round — Problem 30
题目详情
- [10] The sequence a , a , a , . . . of real numbers satisfies the recurrence
1 2 3
a − a + 2 a 2
n − 1 nna = .
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
英文原题
[10] The sequence a 1 , a 2 , a 3 , . . . of real numbers satisfies the recurrence
a n +1 = a 2
n − a n − 1 + 2 a n
a n − 1 + 1 .
Given that a 1 = 1 and a 9 = 7, find a 5 .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
HARVARD-MIT MATHEMATICS TOURNAMENT, MARCH 15, 2003 — GUTS ROUND
解析
英文解析
- The sequence a , a , a , . . . of real numbers satisfies the recurrence
1 2 3
a − a + 2 a 2
n − 1 nna = .
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 nso b = b /b . It follows that the sequence ( b ) is a geometric progression, from 2
n +1 n − 1 nnwhich b = b b = 2 · 8 = 16 ⇒ b = ± 4. However, since all b are real, they either 2
1 9 5 nalternate in sign or all have the same sign (depending on the sign of the progression’s 5
common ratio); either way, b has the same sign as b , so b = 4 ⇒ a = 3.
5 1 5 5