HMMT 二月 2016 · 代数 · 第 5 题
HMMT February 2016 — Algebra — Problem 5
题目详情
英文原题
- An infinite sequence of real numbers a , a , . . . satisfies the recurrence
1 2
a = a − 2 a + an +3 n +2 n +1 nfor every positive integer n . Given that a = a = 1 and a = a , compute a + a + · · · + a .
1 3 98 99 1 2 100
解析
英文解析
- An infinite sequence of real numbers a , a , . . . satisfies the recurrence
1 2
a = a − 2 a + an +3 n +2 n +1 nfor every positive integer n . Given that a = a = 1 and a = a , compute a + a + · · · + a .
1 3 98 99 1 2 100
Proposed by: Evan Chen
Answer: 3
A quick telescope gives that a + · · · + a = 2 a + a + a − a for all n ≥ 3:
1 n 1 3 n − 1 n − 2
n n − 3
∑ ∑
a = a + a + a + ( a − 2 a + 2 a )
k 1 2 3 k k +1 k +2
k =1 k =1
n − 3 n − 2 n − 1
∑ ∑ ∑
= a + a + a + a − 2 a + a
1 2 3 k k kk =1 k =2 k =3 = 2 a + a − a + a .
1 3 n − 2 n − 1
Putting n = 100 gives the answer.
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One actual value of a which yields the sequence is a = .
2 2
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