ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 428

\[1)\ \frac{P_{n}}{P_{n + 1}} = \frac{1}{3}\]

\[\frac{n!}{(n + 1)!} = \frac{1}{3}\]

\[\frac{n!}{(n + 1)n!} = \frac{1}{3}\]

\[\frac{1}{n + 1} = \frac{1}{3}\]

\[n + 1 = 3\]

\[n = 2.\]

\[Ответ:\ \ 2.\]

\[2)\ \frac{nP_{n - 2}}{P_{n}} = 0,1\]

\[\frac{n(n - 2)!}{n!} = 0,1\]

\[\frac{n(n - 2)!}{n(n - 1)(n - 2)!} = 0,1\]

\[\frac{1}{n - 1} = \frac{1}{10}\]

\[n - 1 = 10\]

\[n = 11.\]

\[Ответ:\ \ 11.\]

\[3)\ \frac{2P_{n - 1}}{P_{n + 1}} - 1 = 0\]

\[\frac{2(n - 1)!}{(n + 1)!} = 1\]

\[\frac{2(n - 1)!}{(n + 1)n(n - 1)!} = 1\]

\[\frac{2}{n(n + 1)} = 1\]

\[\frac{1}{n^{2} + n} = \frac{1}{2}\]

\[n^{2} + n = 2\]

\[n^{2} + n - 2 = 0\]

\[D = 1 + 8 = 9\]

\[n_{1} = \frac{- 1 - 3}{2} = - 2;\text{\ \ }\]

\[n_{2} = \frac{- 1 + 3}{2} = 1.\]

\[Ответ:\ \ корней\ нет.\]