Найдите сумму первых шести членов геометрической прогрессии: корень из 2; 2; ….

\[S_{5} = \frac{b_{1}\left( q^{5} - 1 \right)}{q - 1}\]

\[\sqrt{2};2;\ldots\ \]

\[b_{1} = \sqrt{2};\ \ q = \sqrt{2}:\]

\[S_{5} = \frac{\sqrt{2}\left( \left( \sqrt{2} \right)^{5} - 1 \right)}{\sqrt{2} - 1} =\]

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

\[= \frac{\left( 8 - \sqrt{2} \right)\left( \sqrt{2} + 1 \right)}{\left( \sqrt{2} - 1 \right)\left( \sqrt{2} + 1 \right)} =\]

\[= 8\sqrt{2} - 2 + 8 - \sqrt{2} = 7\sqrt{2} + 6.\]