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HMMT 二月 2006 · 代数 · 第 9 题

HMMT February 2006 — Algebra — Problem 9

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

  1. Compute the value of the infinite series

    4 2

    n + 3 n + 10 n + 10
    n 4
    2 · ( n + 4)
    n =2
解析

英文解析

  1. Compute the value of the infinite series

    4 2

    n + 3 n + 10 n + 10
    n 4
    2 · ( n + 4)
    n =2
    Answer:113
    Solution: We employ the difference of squares identity, uncovering the factorization 10
    4 2 2 2 2 2
    of the denominator: n + 4 = ( n + 2) − (2 n ) = ( n − 2 n + 2)( n + 2 n + 2). Now,
    4 2 2
    n + 3 n + 10 n + 10 3 n + 10 n + 6 = 1 +
    4 4
    n + 4 n + 4
    4 1 = 1 + −
    2 2
    n − 2 n + 2 n + 2 n + 2
    ∞ ∞
    4 2
    ∑ ∑
    n + 3 n + 10 n + 10 1 4 1 = ⇒ = + −
    n 4 n n 2 n 2
    2 · ( n + 4) 2 2 · ( n − 2 n + 2) 2 · ( n + 2 n + 2)
    n =2 n =2


    1 1 1 = + −
    n − 2 2 n 2
    2 2 · (( n − 1) + 1) 2 · (( n + 1) + 1)
    n =2
    1 1 1 1 1 11
    The last series telescopes to + , which leads to an answer of + + = .
    2 10 2 2 10 10