Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 807

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

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

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

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

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

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

\[D = 1 + 8 = 9\]

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

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

\[Ответ:\ \ корней\ нет.\]

\[2)\ A_{n + 1}^{2} = 156\]

\[\frac{(n + 1)!}{(n + 1 - 2)!} = 156\]

\[(n + 1)! = 156(n - 1)!\]

\[(n + 1)n(n - 1)! = 156(n - 1)!\]

\[(n + 1)n = 156\]

\[n^{2} + n - 156 = 0\]

\[D = 1 + 624 = 625\]

\[n_{1} = \frac{- 1 - 25}{2} = - 13;\]

\[n_{2} = \frac{- 1 + 25}{2} = 12.\]

\[Ответ:\ \ 12.\]

\[3)\ C_{n}^{3} = \frac{4}{15}C_{n + 2}^{4}\]

\[\frac{n!}{3!(n - 3)!} = \frac{4}{15} \bullet \frac{(n + 2)!}{4!(n + 2 - 4)!}\]

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

\[= \frac{4}{15} \bullet \frac{(n + 2)!}{4 \bullet 3 \bullet 2 \bullet (n - 2)!}\]

\[\frac{n!}{(n - 3)!} = \frac{(n + 2)(n + 1)n!}{15(n - 2)(n - 3)!}\]

\[15(n - 2) = (n + 2)(n + 1)\]

\[15n - 30 = n^{2} + n + 2n + 2\]

\[n^{2} - 12n + 32 = 0\]

\[D = 144 - 128 = 16\]

\[n_{1} = \frac{12 - 4}{2} = 4;\]

\[n_{2} = \frac{12 + 4}{2} = 8.\]

\[Ответ:\ \ 4;\ 8.\]

\[4)\ 12C_{n + 3}^{n - 1} = 5A_{n + 1}^{2}\]

\[12 \bullet \frac{(n + 3)!}{(n - 1)!(n + 3 - n + 1)!} =\]

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

\[\frac{12(n + 3)!}{(n - 1)! \bullet 4!} = \frac{5(n + 1)!}{(n - 1)!}\]

\[\frac{12(n + 3)(n + 2)(n + 1)!}{4 \bullet 3 \bullet 2} =\]

\[= 5(n + 1)!\]

\[(n + 3)(n + 2) = 5 \bullet 2\]

\[n^{2} + 2n + 3n + 6 = 10\]

\[n^{2} + 5n - 4 = 0\]

\[D = 25 + 16 = 41\]

\[\sqrt{D} \notin Q;\ n \in \varnothing.\]

\[Ответ:\ \ корней\ нет.\]