Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1067

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\[1)\ \frac{P_{n}}{P_{n + 1}} = \frac{1}{4}\]

\[\frac{n!}{(n + 1)!} = \frac{1}{4}\]

\[\frac{n!}{(n + 1) \bullet n!} = \frac{1}{4}\]

\[\frac{1}{n + 1} = \frac{1}{4}\]

\[n + 1 = 4\ \]

\[n = 3\]

\[Ответ:\ \ n = 3.\]

\[2)\ \frac{P_{n + 2}}{P_{n + 1}} = 5\]

\[\frac{(n + 2)!}{(n + 1)!} = 5\]

\[\frac{(n + 2) \bullet (n + 1)!}{(n + 1)!} = 5\]

\[n + 2 = 5\]

\[n = 3\]

\[Ответ:\ \ n = 3.\]

\[3)\ \frac{P_{n}}{P_{n - 2}} = 20\]

\[\frac{n!}{(n - 2)!} = 20\]

\[\frac{n \bullet (n - 1) \bullet (n - 2)!}{(n - 2)!} = 20\]

\[n^{2} - n - 20 = 0\]

\[D = 1^{2} + 4 \bullet 20 = 1 + 80 = 81\]

\[n_{1} = \frac{1 - 9}{2} = - 4\ \ и\ \ \]

\[n_{2} = \frac{1 + 9}{2} = 5.\]

\[Ответ:\ \ n = 5.\]

\[4)\ \frac{P_{n - 1}}{P_{n + 1}} = \frac{1}{12}\]

\[\frac{(n - 1)!}{(n + 1)!} = \frac{1}{12}\]

\[\frac{(n - 1)!}{(n + 1) \bullet n \bullet (n - 1)!} = \frac{1}{12}\]

\[\frac{1}{n^{2} + n} = \frac{1}{12}\]

\[n^{2} + n - 12 = 0\]

\[D = 1^{2} + 4 \bullet 12 = 1 + 48 = 49\]

\[n_{1} = \frac{- 1 - 7}{2} = - 4\ \ и\ \ \]

\[n_{2} = \frac{- 1 + 7}{2} = 3.\]

\[Ответ:\ \ n = 3.\]