HMMT 二月 2016 · 代数 · 第 5 题

HMMT February 2016 — Algebra — Problem 5

An infinite sequence of real numbers a , a , . . . satisfies the recurrence 1 2 a = a − 2 a + a n +3 n +2 n +1 n for 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

An infinite sequence of real numbers a , a , . . . satisfies the recurrence 1 2 a = a − 2 a + a n +3 n +2 n +1 n for 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 k k =1 k =2 k =3 = 2 a + a − a + a . 1 3 n − 2 n − 1 Putting n = 100 gives the answer. 742745601954 One actual value of a which yields the sequence is a = . 2 2 597303450449