HMMT 二月 2020 · ALGNT 赛 · 第 8 题
HMMT February 2020 — ALGNT Round — Problem 8
题目详情
- Let P ( x ) be the unique polynomial of degree at most 2020 satisfying P ( k ) = k for k = 0 , 1 , 2 , . . . , 2020. 2 Compute P (2021 ). 2020
解析
- Let P ( x ) be the unique polynomial of degree at most 2020 satisfying P ( k ) = k for k = 0 , 1 , 2 , . . . , 2020. 2 Compute P (2021 ). Proposed by: Milan Haiman ( ) 4040 Answer: 2021 − 2020 2 P ( x ) − x Solution 1: Since P (0) = 0, we see that P has no constant term. Let Q ( x ) = be a polynomial x with degree at most 4039. From the given values of P , we see that Q ( k ) = 0 and Q ( − k ) = − 2 for k = 1, 2, 3, . . . , 2020. Now, consider the polynomial R ( x ) = Q ( x + 1) − Q ( x ), which has degree at most 4038. Then R has roots − 2020, − 2019, . . . , − 2, 1, 2, . . . , 2019, so R ( x ) = a ( x + 2020) · · · ( x + 2)( x − 1) · · · ( x − 2019) 1 for some real number a . Using R (0) + R ( − 1) = Q (1) − Q ( − 1) = 2 yields a = − , so 2020!2019! ( ) 4040 · · · 2022 · 2019 · · · 1 1 4040 Q (2021) = R (2020) + Q (2020) = − + 0 = − . 2020!2019! 2021 2020 ( ) 4040 2 It follows that P (2021 ) = 2021 Q (2021) + 2021 = 2021 − . 2020 Solution 2: By Lagrange interpolation, 2020 2020 2020 2 k ∑ ∏ ∑ ∏ ( x − j ) 2( − 1) k 2 P ( x ) = k = ( x − j ) . 2 2 2 ( k − j ) (2020 − k )!(2020 + k )!( x − k ) 0 ≤ j ≤ 2020 j =0 k =0 k =0 j 6 = k Therefore, by applying Pascal’s identity multiple times, we get that 2020 k ∑ 4042!( − 1) k 2 P (2021 ) = (2021 − k )!(2021 + k )! k =0 ( ) 2020 ∑ 4042 k = ( − 1) k 2021 − k k =0 ( ) (( ) ( )) 2021 ∑ 4041 4041 k +1 = 2021 − ( − 1) k + 2021 − k 2020 − k k =0 ( ) ( ) 2021 ∑ 4041 k +1 = 2021 − ( − 1) 2021 − k k =1 ( ) (( ) ( )) 2021 ∑ 4040 4040 k +1 = 2021 − ( − 1) + 2021 − k 2020 − k k =1 ( ) 4040 = 2021 − . 2020 2020