Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 473

Авторы:
Год:2020-2021-2022-2023
Тип:учебник

Задание 473

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

\[\frac{(n + 2)!}{n!} = 12\]

\[\frac{(n + 2)(n + 1)n!}{n!} = 12\]

\[(n + 2)(n + 1) = 12\]

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

\[n^{2} + 3n - 10 = 0\]

\[D = 9 + 40 = 49\]

\[n_{1} = \frac{- 3 - 7}{2} = - 5;\text{\ \ }\]

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

\[Ответ:\ \ 2.\]

\[2)\ \frac{1}{P_{n - 4}} = \frac{20}{P_{n - 2}}\]

\[\frac{1}{(n - 4)!} = \frac{20}{(n - 2)!}\]

\[\frac{1}{(n - 4)!} = \frac{20}{(n - 2)(n - 3)(n - 4)!}\]

\[\frac{20}{(n - 2)(n - 3)} = 1\]

\[(n - 2)(n - 3) = 20\]

\[n^{2} - 3n - 2n + 6 = 20\]

\[n^{2} - 5n - 14 = 0\]

\[D = 25 + 56 = 81\]

\[n_{1} = \frac{5 - 9}{2} = - 2;\text{\ \ }\]

\[n_{2} = \frac{5 + 9}{2} = 7.\]

\[Ответ:\ \ 7.\]

\[3)\ A_{n + 1}^{4} = 6n(n + 1)\]

\[\frac{(n + 1)!}{(n - 3)!} = 6n(n + 1)\]

\[\frac{(n + 1)n(n - 1)(n - 2)(n - 3)!}{(n - 3)!} =\]

\[= 6n(n + 1)\]

\[(n - 1)(n - 2) = 6\]

\[n^{2} - 2n - n + 2 = 6\]

\[n^{2} - 3n - 4 = 0\]

\[D = 9 + 16 = 25\]

\[n_{1} = \frac{3 - 5}{2} = - 1;\text{\ \ }\]

\[n_{2} = \frac{3 + 5}{2} = 4.\]

\[Ответ:\ \ 4.\]

\[4)\ A_{n - 1}^{5} = 2A_{n - 2}^{5}\]

\[\frac{(n - 1)!}{(n - 6)!} = 2 \bullet \frac{(n - 2)!}{(n - 7)!}\]

\[\frac{(n - 1)(n - 2)!}{(n - 6)(n - 7)!} = 2 \bullet \frac{(n - 2)!}{(n - 7)!}\]

\[\frac{n - 1}{n - 6} = 2\]

\[n - 1 = 2(n - 6)\]

\[n - 1 = 2n - 12\]

\[n = 11.\]

\[Ответ:\ \ 11.\]

\[5)\ \frac{C_{n + 1}^{3}}{C_{n}^{4}} = \frac{8}{5}\]

\[5C_{n + 1}^{3} = 8C_{n}^{4}\]

\[5 \bullet \frac{(n + 1)!}{3!(n - 2)!} = 8 \bullet \frac{n!}{4!(n - 4)!}\]

\[5 \bullet \frac{(n + 1)n!}{3 \bullet 2 \bullet (n - 2)(n - 3)(n - 4)!} =\]

\[= 8 \bullet \frac{n!}{4 \bullet 3 \bullet 2 \bullet (n - 4)!}\]

\[\frac{5(n + 1)}{6(n - 2)(n - 3)} = \frac{8}{24}\]

\[120(n + 1) = 48(n - 2)(n - 3)\]

\[2\left( n^{2} - 3n - 2n + 6 \right) = 5(n + 1)\]

\[2n^{2} - 10n + 12 = 5n + 5\]

\[2n^{2} - 15n + 7 = 0\]

\[D = 225 - 56 = 169\]

\[n_{1} = \frac{15 - 13}{2 \bullet 2} = \frac{1}{2};\ \]

\[n_{2} = \frac{15 + 13}{2 \bullet 2} = 7.\]

\[Ответ:\ \ 7.\]

\[6)\ C_{n}^{3} = 4C_{n - 2}^{2}\]

\[\frac{n!}{3!(n - 3)!} = 4 \bullet \frac{(n - 2)!}{2!(n - 4)!}\]

\[\frac{n(n - 1)(n - 2)!}{3 \bullet 2 \bullet (n - 3)(n - 4)!} =\]

\[= 4 \bullet \frac{(n - 2)!}{2 \bullet (n - 4)!}\]

\[\frac{n(n - 1)}{6(n - 3)} = 2\]

\[n(n - 1) = 12(n - 3)\]

\[n^{2} - n = 12n - 36\]

\[n^{2} - 13n + 36 = 0\]

\[D = 169 - 144 = 25\]

\[n_{1} = \frac{13 - 5}{2} = 4;\ \]

\[n_{2} = \frac{13 + 5}{2} = 9.\]

\[Ответ:\ \ 4;\ 9.\]

Скачать ответ
Есть ошибка? Сообщи нам!

Решебники по другим предметам