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HMMT 二月 2006 · 代数 · 第 4 题

HMMT February 2006 — Algebra — Problem 4

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

  1. Let a , a , . . . be a sequence defined by a = a = 1 and a = a + a for n ≥ 1. Find
    1 2 1 2 n +2 n +1 n


    .nan +1
    n =14
解析

英文解析

  1. Let a , a , . . . be a sequence defined by a = a = 1 and a = a + a for n ≥ 1.
    1 2 1 2 n +2 n +1 n
    Find


    .nan +1
    n =14
    Answer:1
    Solution: Let X denote the desired sum. Note that 11
    1 1 2 3 5
    X = + + + + + . . .
    2 3 4 5 6
    4 4 4 4 4
    1 1 2 3 5 8
    4 X = + + + + + + . . .
    1 2 3 4 5 6
    4 4 4 4 4 4
    1 1 2 3 5 8 13
    16 X = + + + + + + + . . .
    0 1 2 3 4 5 6
    4 4 4 4 4 4 4
    so that X + 4 X = 16 X − 1, and X = 1 / 11.