Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.4 AIME, harder Prove it Taiwan

對任意自然數 $n$ , 設 $a_n = \sum_{k=1}^{\infty} \left[ \frac{n+2^{k-1}}{2^k} \right]$ , 其中 $[x]$ 表示不超過 $x$ 的最大整數。試求 $a_{2015}$ 之值。

For any positive integer $n$ , let $a_n = \sum_{k=1}^{\infty} \left[ \frac{n+2^{k-1}}{2^k} \right]$ , where $[x]$ is the largest integer that is equal or less than $x$ . Determine the value of $a_{2015}$ .

Solution

By case analysis it is easy to see that
$\left[ x + \frac{1}{2} \right] = [2x] - [x]$
holds for any real number $x$ , and therefore
$\left[ \frac{n + 2^{k-1}}{2^k} \right] = \left[ \frac{n}{2^{k-1}} \right] - \left[ \frac{n}{2^k} \right],$
from which we obtain
$\sum_{k=1}^{L} \left[ \frac{n + 2^{k-1}}{2^k} \right] = [n] - \left[ \frac{n}{2^L} \right],$
hence $a_n = [n]$ , and thus $a_{2015} = 2015$ .

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from zh; metadata (topic, difficulty) added by this project.