Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 433

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\[S = \frac{16}{3};\ \ b_{n} = \frac{1}{6};\ \frac{S_{n - 1}}{S_{n + 1}} = 30.\]

\[S_{n - 1} = 30 \cdot S_{n + 1} = 30 \cdot \frac{b_{n + 1}}{1 - q};\]

\[S = S_{n - 1} + \frac{1}{6} + S_{n + 1} =\]

\[= 30 \cdot S_{n + 1} + \frac{1}{6} + S_{n + 1} =\]

\[= \frac{1}{6} + 31 \cdot S_{n + 1};\]

\[31S_{n + 1} = \frac{16}{3} - \frac{1}{6} = \frac{31}{6}\]

\[S_{n + 1} = \frac{1}{6}.\]

\[\frac{b_{n + 1}}{1 - q} = \frac{1}{6}:\]

\[\frac{S_{n + 1}}{S} = \frac{b_{n + 1}}{1 - q}\ :\frac{b_{1}}{1 - q} =\]

\[= \frac{b_{n + 1}}{b_{1}} = q^{n};\]

\[\frac{1}{6}\ :\frac{16}{3} = q^{n}\]

\[\frac{1}{32} = q^{n}\]

\[q^{n} = \left( \frac{1}{2} \right)^{5}\]

\[n = 5.\]

\[Ответ:n = 5.\]