Question

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

Answer 1

$\{b_{n}\}:\; \; -4\sqrt2\; ,\; 4\; ,\; -2\sqrt2\; ,\; ...\\\\q=\frac{b_{n}}{b_{n-1}}\; \; ,\qquad \frac{b_2}{b_1}=\frac{4}{-4\sqrt2}=-\frac{1}{\sqrt2} \; \; ,\; \; \; \frac{b_3}{b_2}=\frac{-2\sqrt2}{4}=-\frac{\sqrt2}{2}=-\frac{1}{\sqrt2}\\\\\frac{b_2}{b_1}=\frac{b_3}{b_2}\; \; \Rightarrow \; \; \; q=-\frac{1}{\sqrt2}\\\\S=\frac{b_1}{1-q}=\frac{-4\sqrt2}{1+\frac{1}{\sqrt2}}=\frac{-4\sqrt2\, \cdot \, \sqrt2}{\sqrt2+1}=-\frac{8}{\sqrt2+1}=-\frac{8\, (\sqrt2-1)}{(\sqrt2+1)(\sqrt2-1)}=-8\, (\sqrt2-1)=$

$=8\, (1-\sqrt2)$