Question

Найдите сумму первых шести членов геометрической прогрессии (b↓n), в которой: b↓1=3√2, q=√2

Answer 1

$\displaystyle\bf\\b_{1} =3 \ \ ; \ \ q=\sqrt{2}\\\\\\S_{6}=\frac{b_{1} \cdot(q^{6} -1)}{q-1} =\frac{3\cdot\Big[(\sqrt{2})^{6} -1\Big] }{\sqrt{2} -1} =\frac{3\cdot(8-1)}{\sqrt{2} -1} =\frac{21}{ \sqrt{2}-1 } =\\\\\\=\frac{21\cdot(\sqrt{2} +1)}{(\sqrt{2} -1)(\sqrt{2} +1)} =\frac{21\cdot(\sqrt{2} +1)}{(\sqrt{2}) ^{2} -1^{2} } =\frac{21\cdot(\sqrt{2} +1)}{2-1} =21(\sqrt{2}+1)$