PUMaC 2019 · 代数（A 组） · 第 7 题
PUMaC 2019 — Algebra (Division A) — Problem 7
题目详情
- A doubly-indexed sequence a , for m and n nonnegative integers, is defined as follows. m,n (a) a = 0 for all m > 0 and a = 1. m, 0 0 , 0 (b) a = 0 for all m > 1, and a = 1 , a = 0. m, 1 1 , 1 0 , 1 (c) a = a + a for all n ≥ 2 0 ,n 0 ,n − 1 0 ,n − 2 (d) a = a + a + a − a for all m > 0 , n ≥ 2. m,n m,n − 1 m,n − 2 m − 1 ,n − 1 m − 1 ,n − 2 ∞ ∞ m ∑ ∑ a x m,n 2 Then there exists a unique value of x so = 1. Find b 1000 x c . n − m 3 m =0 n =0
解析
- A doubly-indexed sequence a , for m and n nonnegative integers, is defined as follows. m,n (a) a = 0 for all m > 0 and a = 1. m, 0 0 , 0 (b) a = 0 for all m > 1, and a = 1 , a = 0. m, 1 1 , 1 0 , 1 (c) a = a + a for all n ≥ 2 0 ,n 0 ,n − 1 0 ,n − 2 (d) a = a + a + a − a for all m > 0 , n ≥ 2. m,n m,n − 1 m,n − 2 m − 1 ,n − 1 m − 1 ,n − 2 ∞ ∞ m ∑ ∑ a x m,n 2 Then there exists a unique value of x so = 1. Find b 1000 x c . n − m 3 m =0 n =0 Proposed by: Frank Lu Answer: 27 Define the sequence of polynomials P ( x ) by P ( x ) = 1 , P ( x ) = x , and for n ≥ 2 P ( x ) = n 0 1 n ( x + 1) P ( x ) − ( x − 1) P ( x ). Observe that our given sequence is uniquely determined n − 1 n − 2 m by the values when n = 0 and n = 1, over all m . Letting b be the coefficient of x in m,n P ( x ), our recurrence becomes b = b + b + b − b , and that the n m,n m,n − 1 m,n − 2 m − 1 ,n − 1 m − 1 ,n − 2 b satisfy the same initial conditions that are given. It thus follows that our b sequence m,n ∞ ∑ n is the exact same as the given a sequence. Now, define the function g ( x, y ) = P ( x ) y . n n =0 Observe that, based on our recurrence and the initial conditions, we have that g ( x, y ) = 2 1 + xy + ( x + 1) y ( g ( x, y ) − 1) − ( x − 1) y ( g ( x, y )). Rearranging this and solving for g gives us 1 − y now that g ( x, y ) = . But our desired sum can be written as g (3 x, 1 / 3). Let 2 y ( x − 1) − y ( x +1)+1 2 2 − 2 u 5 − 1 3 u = 3 x . Thus, our desired u is so that = 1, or that = + , or u = , 1 1 3 9 9 2 ( u − 1) − ( u +1)+1 9 3 − 1 1000 250 which means that x = . This gives us an answer of b c = b c = 27 6 36 9