# Expanding the sum of two natural valued terms.

I was looking at the IMO 2013 problems and I was trying to solve the first problem.

Prove that for any pair of positive integers $k$ and $n$, there exist $k$ positive integers $m_1$,$m_2$,$...$,$m_k$ (not necessarily different) such that: $$1+\frac{2^k-1}{n}=\left(1+\frac{1}{m_1}\right)\left(1+\frac{1}{m_2}\right)...\left(1+\frac{1}{m_k}\right)$$ However I got stuck and I was very frustrated so I decided to look up for the solution. The solution follows my logic (I was going in the right direction) but it states that:

$$1+\frac{2^{m+1}-1}{n}=\left( 1+\frac{2^{m}-1}{\frac{n}{2}}\right)\cdot\left( 1+\frac{1}{n+2^{m+1}-2}\right)$$ Suposing that $n$ is arbitrary fixed and even; and $m\in\mathbb{Z}^{+}$.

How is the expansion performed?

• Sorry, my mistake. The question is how do I perform such expansion?
– avm
May 25 '14 at 22:39

Start by multiplying numerator and denominator by $\frac{1}{2} .$ \begin{align} 1 + \frac{2^m - 1}{n/2} + \frac{1}{n} &= 1 + \frac{2^m - 1}{n/2} + \frac{n + 2^{m+1} - 2}{n(n + 2^{m+1} - 2)} \\ &= 1 + \frac{2^{m+1} - 2}{n} + \frac{1}{n + 2^{m+1} - 2} + \frac{2^{m+1} - 2}{n(n + 2^{m+1} - 2)} \\ &= 1 + \frac{2^{m+1} - 2}{n} + \frac{1}{n + 2^{m+1} - 2} + \frac{2^{m+1} - 2}{n}\frac{1}{(n + 2^{m+1} - 2)}\\ \end{align}

Why we must replace $1$ by the term $\frac{n + 2^{m+1} - 2}{(n + 2^{m+1} - 2)}$ is still not intuitive, but it gives the result.