Найдите сумму четырёх первых членов геометрической прогрессии (bn) со знаменателем q, если: b1=2*корень из 2, b7=16*корень из 2, q>0.

\[b_{1} = 2\sqrt{2};\ \ \ \ \ \ \ \ b_{7} = 16\sqrt{2};\ \ \ \ \ \ \ \]

\[q > 0\]

\[b_{1}q^{6} = 16\sqrt{2}\text{\ \ \ \ \ \ }\]

\[q^{6} = \frac{16\sqrt{2}}{2\sqrt{2}} = 8\ \ \ \ \]

\[q = \sqrt{2}.\]

\[S_{4} = \frac{2\sqrt{2} \cdot \left( \left( \sqrt{2} \right)^{4} - 1 \right)}{\sqrt{2} - 1} =\]

\[= \frac{2\sqrt{2} \cdot (4 - 1)}{\sqrt{2} - 1} =\]

\[= \frac{6\sqrt{2}\left( \sqrt{2} + 1 \right)}{\left( \sqrt{2} - 1 \right)\left( \sqrt{2} + 1 \right)} =\]

\[= \frac{6 \cdot 2 + 6\sqrt{2}}{2 - 1} = 12 + 6\sqrt{2}.\]

\[Ответ:12 + 6\sqrt{2}.\]