HMMT 二月 2004 · 代数 · 第 4 题
HMMT February 2004 — Algebra — Problem 4
题目详情
英文原题
- Evaluate the sum
1 1 1 1
√ √ √ √
-
-
- · · · + .
2 b 1 c + 1 2 b 2 c + 1 2 b 3 c + 1 2 b 100 c + 1 − 1
- · · · + .
-
解析
英文解析
- Evaluate the sum
1 1 1 1
√ √ √ √
-
-
- · · · + .
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 theanswer is 9 + 1 / 21 = 190 / 21.
− 1
- · · · + .
-