mathematical olympiad

IMO 2013 解答 Problem 1 (Japan) 证明 1 注意到 \begin{equation}\left(1+\frac1{2a}\right)\left(1+\frac1{2a+1}\right)=1+\frac1a,\end{equation} 于是, 我们可以对形如 \begin{equation}\left(1+\dfrac1b\right)\left(1+\dfrac1{b+1}\right)\dotsm\left(1+\dfrac1{b+2^t-2}\right)\end{equation} 的乘积进行操作: 首先把因式按照如下规则配对: 若 \(b\) 是奇数, \(1+\dfrac1b\) 不动, 把 \(1+\dfrac1{b+1}\) 与 \(1+\dfrac1{b+2}\) 配对, \(\dotsc\), \(1+\dfrac1{b+2^t-3}\) 与 \(1+\dfrac1{b+2^t-2}\) 配对; …

                                       Day \(1\)   Tuesday, July 23, 2013 Problem 1. Prove that for …