PUMaC 2023 · 个人决赛（B 组） · 第 2 题

PUMaC 2023 — Individual Finals (Division B) — Problem 2

Let f be a polynomial with degree at most n − 1. Show that n X n k ( − 1) f ( k ) = 0 . k k =0

解析

Let f be a polynomial with degree at most n − 1. Show that n X n k ( − 1) f ( k ) = 0 . k k =0 Proposed by Ben Zenker, solved by Atharva Pathak Solution: It suffices to show this for the polynomials f ( X ) = X ( X − 1) · · · ( X − ℓ + 1) for 0 ≤ ℓ ≤ n − 1, since all other polynomials of degree at most n − 1 can be written as finite 1 linear combinations of them. We have n X n k ( − 1) · k ( k − 1) · · · ( k − ℓ + 1) k k =0 n X n ! k = ( − 1) · k ( k − 1) · · · ( k − ℓ + 1) k !( n − k )! k = ℓ n X n ! k = ( − 1) ( k − ℓ )!( n − k )! k = ℓ n ℓ X ( − 1) n ! ( n − ℓ )! k − ℓ n − k = ( − 1) 1 ( n − ℓ )! ( k − ℓ )!( n − k )! k = ℓ ℓ ( − 1) n ! n − ℓ = ( − 1 + 1) ( n − ℓ )! = 0 , as desired.