Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 434

\[\boxed{\mathbf{434}.}\]

\[\frac{b_{1}^{2}\left( 1 - q^{2n} \right)}{1 - q^{2}} = \frac{b_{1}\left( 1 - q^{2n} \right)}{1 - q}\text{\ \ \ \ }\]

\[\ | \cdot \frac{1 - q}{b_{1}}\]

\[\frac{b_{1}}{1 + q} = 1.\]

\[\frac{b_{1}^{3}\left( 1 - q^{3n} \right)}{1 - q^{3}}\ :\frac{b_{1}\left( 1 - q^{3n} \right)}{1 - q} = \frac{1}{3}\]

\[\frac{b_{1}^{3}\left( 1 - q^{3n} \right)(1 - q)}{\left( 1 - q^{3} \right)b_{1}\left( 1 - q^{3n} \right)} = \frac{1}{3}\]

\[\frac{b_{1}^{2}}{1 + q + q^{2}} = \frac{1}{3}\]

\[b_{1} = 1 + q:\]

\[\frac{(1 + q)^{2}}{1 + q + q^{2}} - \frac{1}{3} = 0\]

\[2q^{2} - 5q + 2 = 0\]

\[D = 25 - 16 = 9\]

\[q_{1} = \frac{- 5 + 3}{4} = - \frac{1}{2};\ \ \]

\[q_{2} = \frac{- 5 - 3}{4} = - 2.\]

\[q = - 0,5:\]

\[b_{1} = 1 - 0,5 = 0,5.\]

\[S = \frac{0,5}{1 + 0,5} = \frac{5}{15} = \frac{1}{3}.\]

\[Ответ:\ \frac{1}{3}.\]