\(1.3.5.7.9...59=\frac{\left(1.3.5...59\right).\left(2.4.6...60\right)}{2.4.6...60}=\frac{1.2.3...60}{2^{30}\left(1.2.3...30\right)}\)
\(=\frac{31.32.33...60}{2.2.2...2}=\frac{31}{2}\cdot\frac{32}{2}\cdot\frac{33}{2}...\frac{60}{2}\)
Vậy \(\frac{31}{2}\cdot\frac{32}{2}\cdot\frac{33}{2}...\frac{60}{2}=1.3.5...59\)(đpcm)