HMMT 二月 2004 · 代数 · 第 4 题
HMMT February 2004 — Algebra — Problem 4
题目详情
- Evaluate the sum 1 1 1 1 √ √ √ √
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- · · · + . 2 b 1 c + 1 2 b 2 c + 1 2 b 3 c + 1 2 b 100 c + 1 − 1
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解析
- Evaluate the sum 1 1 1 1 √ √ √ √
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- · · · + . 2 b 1 c + 1 2 b 2 c + 1 2 b 3 c + 1 2 b 100 c + 1 Solution: 190 / 21 The first three terms all equal 1 / 3, then the next five all equal 1 / 5; more generally, ⌊ ⌋ ⌊ ⌋ √ √ 2 2 for each a = 1 , 2 , . . . , 9, the terms 1 / (2 a + 1) to 1 / (2 a + 2 a + 1) all equal 1 / (2 a + 1), and there are 2 a + 1 such terms. Thus our terms can be arranged into ⌊ ⌋ √ 9 groups, each with sum 1, and only the last term 1 / (2 100 + 1) remains, so the answer is 9 + 1 / 21 = 190 / 21. − 1
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