pierwiastek

[Filip25]

Wykaż, że \( \sqrt{2 \sqrt{7 \sqrt{2 \sqrt{7...} } } } = \sqrt[3]{28} \)

[Icanseepeace]

\( \sqrt{2 \sqrt{7 \sqrt{2 \sqrt{7...} } } } = 2^{\frac{1}{2} + \frac{1}{8} + \frac{1}{32} + \ldots} \cdot 7^{\frac{1}{4} + \frac{1}{16} + \ldots} = 2^{\frac{1}{2} \cdot \frac{1}{1 - \frac{1}{4}}} \cdot 7^{\frac{1}{4} \cdot \frac{1}{1 - \frac{1}{4}}} = \sqrt[3]{28} \)