Let n\in\mathbb{N}. Prove that:
\sum_{k = 0}^{m}\frac{{2n - k\choose k}}{{2n - k\choose n}}\cdot \frac{2n - 4k + 1}{2n - 2k + 1}\cdot 2^{n - 2k} = \frac{{n\choose m}}{{2n - 2m\choose n - m}}\cdot 2^{n - 2m}