HMMT 二月 2017 · 冲刺赛 · 第 28 题

HMMT February 2017 — Guts Round — Problem 28

[ 15 ] Let . . . , a , a , a , a , . . . be a sequence of positive integers satisfying the following relations: − 1 0 1 2 a = 0 for n < 0, a = 1, and for n ≥ 1, n 0 a = a + 2( n − 1) a + 9( n − 1)( n − 2) a + 8( n − 1)( n − 2)( n − 3) a . n n − 1 n − 2 n − 3 n − 4 Compute n ∑ 10 a n . n ! n ≥ 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . February 2017, February 18, 2017 — GUTS ROUND Organization Team Team ID#

解析

[ 15 ] Let . . . , a , a , a , a , . . . be a sequence of positive integers satisfying the following relations: − 1 0 1 2 a = 0 for n < 0, a = 1, and for n ≥ 1, n 0 a = a + 2( n − 1) a + 9( n − 1)( n − 2) a + 8( n − 1)( n − 2)( n − 3) a . n n − 1 n − 2 n − 3 n − 4 Compute n ∑ 10 a n . n ! n ≥ 0 Proposed by: Yang Liu 23110 Answer: e . ∑ n x a ′ 2 3 2 3 4 n Let y = . Then y = (1 + 2 x + 9 x + 8 x ) y by definition. So y = C exp( x + x + 3 x + 2 x ) . n ≥ 0 n ! Take x = 0 to get C = 1 . Take x = 10 to get the answer.