Question

5. Найдите сумму бесконечно убывающей геометрической прогрессии:
1) 6, 4, 8/3, ...; 2) 5, -1, 1/5, ..., 3) 1, -1/4, 1/16, ...​

Answer 1

$\boxed {\; S=\dfrac{b_1}{1-q}\; \; ,\; esli\; \; |q|<1\ ;}$

$1)\; \; \{b_{n}\}:\; \; 6\; ,\; 4\; ,\; \dfrac{8}{3}\; ,\; ...\\\\q=\dfrac{b_{n+1}}{b_{n}}\; \; \to \; \; q=\dfrac{b_2}{b_1}=\dfrac{4}{6}=\dfrac{2}{3}\; \; ,\; \; \; \; |q|=\dfrac{2}{3}<1\; \; \; \Rightarrow \\\\\\S=\dfrac{b_1}{1-q}=\dfrac{6}{1-\frac{2}{3}}=\dfrac{6}{\frac{1}{3}}=6\cdot 3=18$

$2)\; \; \{b_{n}\}:\; \; 5\; ,\; -1\; ,\; \dfrac{1}{5}\; ,\; ...\\\\q=\dfrac{b_{n+1}}{b_{n}}\; \; \to \; \; q=\dfrac{b_2}{b_1}=\dfrac{-1}{5}\; \; ,\; \; \; \; |q|=\Big|-\dfrac{1}{5}\Big|=\dfrac{1}{5}<1\; \; \; \Rightarrow \\\\\\S=\dfrac{b_1}{1-q}=\dfrac{5}{1+\frac{1}{5}}=\dfrac{5}{\frac{6}{5}}=\dfrac{5\cdot 5}{6}=\dfrac{25}{6}=4\dfrac{1}{6}$

$3)\; \; \{b_{n}\}:\; \; 1\; ,\; -\dfrac{1}{4}\; ,\; \dfrac{1}{16}\; ,\; ...\\\\q=\dfrac{b_{n+1}}{b_{n}}\; \; \to \; \; q=\dfrac{b_2}{b_1}=\dfrac{-\frac{1}{4}}{1}=-\dfrac{1}{4}\; \; ,\; \; \; \; |q|=\Big|-\dfrac{1}{4}\Big|=\dfrac{1}{4}<1\; \; \; \Rightarrow \\\\\\S=\dfrac{b_1}{1-q}=\dfrac{1}{1+\frac{1}{4}}=\dfrac{1}{\frac{5}{4}}=\dfrac{4}{5}=0,8$