PUMaC 2011 · 代数（A 组） · 第 6 题

PUMaC 2011 — Algebra (Division A) — Problem 6

[ 6 ] A sequence of real numbers { a } ( n = 1 , 2 , ... ) has the following property: n n =1 6 a + 5 a = 20 + 11 a (for n ≥ 3) . n n − 2 n − 1 The first two elements are a = 0 , a = 1. Find the integer closest to a . 1 2 2011 6

解析

Rewrite the equation as a − a = ( a − a ) + . Define another sequence { b } such n n − 1 n − 1 n − 2 n 6 3 5 10 that b = a − a . Thus, b = 1 and b = b + for n ≥ 2, and if we define { c } such n n +1 n 1 n n − 1 n 6 3 2 5 that c = b − 20, then c = − 19 and c = c for n ≥ 2. Now n n 1 n n − 1 6 2010 2010 2010 2010 ∑ ∑ ∑ ∑ a = a + ( a − a ) = b = ( c + 20) = 2010 · 20 + c 2011 0 n n − 1 n n n n =1 n =1 n =1 n =1 ( ) ( ) 2010 5 − 19 · 1 − 6 = + 40200 ≈ 40086 . 5 1 − 6 iπ/ 3 i