Question

Найдите сумму бесконечной геометрической прогрессии -4√2; 4; -2√2

Answer 1

$b_{1}=-4\sqrt{2}\\\\b_{2}=4\\\\b_{2}=b_{1} *q\\\\q=\frac{b_{2}}{b_{1}}=\frac{4}{-4\sqrt{2}}=-\frac{1}{\sqrt{2}}\\\\S=\frac{b_{1}}{1-q}=\frac{-4\sqrt{2}}{1-\frac{1}{\sqrt{2}}}=-\frac{4\sqrt{2}}{\frac{\sqrt{2}-1}{\sqrt{2}}}=-\frac{4\sqrt{2}*\sqrt{2}}{\sqrt{2}-1}=-\frac{8}{\sqrt{2}-1}=-\frac{8*(\sqrt{2}+1)}{(\sqrt{2}-1)(\sqrt{2}+1)}=-\frac{8(\sqrt{2}+1)}{2-1}=-8(\sqrt{2}+1)$