返回题库

HMMT 二月 2009 · 代数 · 第 7 题

HMMT February 2009 — Algebra — Problem 7

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

  1. [ 5 ] Simplify the product
    100 100
    n + m n + m +2 2 n +1 2 m +1
    ∏ ∏
    x + x + x + x
    .
    2 n n + m 2 mx + 2 x + xm =1 n =1
    Express your answer in terms of x .
    2 2 3 3
解析

英文解析

  1. [ 5 ] Simplify the product
    100 100
    n + m n + m +2 2 n +1 2 m +1
    ∏ ∏
    x + x + x + x
    .
    2 n n + m 2 mx + 2 x + xm =1 n =1
    Express your answer in terms of x .
    ( )
    1002
    1+ x 1 1 1
    9900 9900 10000 10100
    Answer: x (OR x + x + x )
    2 4 2 4
    Solution: We notice that the numerator and denominator of each term factors, so the product isequal to
    100 100
    m n +1 m +1 n
    ∏ ∏
    ( x + x )( x + x )
    .
    m n 2
    ( x + x )
    m =1 n =1
    Each term of the numerator cancels with a term of the denominator except for those of the formm 101 101 n
    ( x + x ) and ( x + x ) for m, n = 1 , . . . , 100, and the terms in the denominator which remain are
    1 n 1 mof the form ( x + x ) and ( x + x ) for m, n = 1 , . . . , 100. Thus the product simplifies to
    ( )
    1002
    m 101
    ∏
    x + x
    1 mx + xm =1
    Reversing the order of the factors of the numerator, we find this is equal to
    ( ) ( )
    2 2
    100 100
    101 − m 101 1 m +1
    ∏ ∏
    x + x x + x
    100 − m = x
    1 m 1 mx + x x + xm =1 m =1
    ( )
    1002
    1 1
    ∏
    x + x 01
    100 − m = x
    1 1
    x + xm =1
    ( )
    1002
    99 · 100 1 + x = ( x )22
    as desired.2
    2 2 3 3