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HMMT 十一月 2010 · 冲刺赛 · 第 24 题

HMMT November 2010 — Guts Round — Problem 24

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

  1. [ 12 ] Let P ( x ) be a polynomial of degree at most 3 such that P ( x ) = for x = 1 , 2 , 3 , 4. What is
    1+ x + x 2
    P (5)?
解析

英文解析

  1. [ 12 ] Let P ( x ) be a polynomial of degree at most 3 such that P ( x ) = for x = 1 , 2 , 3 , 4. What is
    1+ x + x 2
    P (5)?
    Guts Round − 3
    Answer: The forward difference of a polynomial P is ∆ P ( x ) = P ( x + 1) − P ( x ), which is a newpolynomial with degree reduced by one. Therefore, if we apply this operation three times we’ll get a 91
    constant function, and we can work back up to get a value of P (5). Practically, we create the followingtable of differences:
    1 1 1 1
    3 7 13 21 − 4 − 6 − 8
    21 91 273
    34 10
    273 273 − 24
    Then extend it to be the following table:273
    1 1 1 1 − 9
    3 7 13 21 273 − 4 − 6 − 8 − 22
    21 91 273 273
    34 10 − 14
    273 273 273 − 24 − 24
    273 273 − 9 − 3
    So our answer is =
    273 91