Решебник по алгебре 9 класс Макарычев Задание 671

 

\[\boxed{\text{671.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ a_{1} = 5,\ \ a_{2} = 10,\ \ q = 5:\]

\[x_{n} = 5 + 5 \cdot (n - 1) = 5 + 5n - 5 = 5n.\ \]

\[\textbf{б)}\ \ a_{1} = 6,\ \ a_{2} = 11,\ \ q = 5:\]

\[x_{n} = 6 + 5 \cdot (n - 1) = 6 + 5n - 5 = 5n + 1.\]

\[\boxed{\text{671.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ x_{n} = 2^{n}\]

\[x_{1} = 2^{1} = 2\]

\[x_{2} = 2^{2} = 4\]

\[x_{3} = 2^{3} = 8\]

\[q = \frac{x_{3}}{x_{2}} = \frac{x_{2}}{x_{1}} = 2 \Longrightarrow\]

\[\Longrightarrow геометрическая\ \]

\[прогрессия.\]

\[\textbf{б)}\ x_{n} = 3^{- n}\]

\[x_{1} = 3^{- 1} = \frac{1}{3}\]

\[x_{2} = 3^{- 2} = \frac{1}{9}\]

\[x_{3} = 3^{- 3} = \frac{1}{27}\]

\[q = \frac{x_{3}}{x_{2}} = \frac{x_{2}}{x_{1}} = \frac{1}{3} \Longrightarrow\]

\[\Longrightarrow геометрическая\ \]

\[прогрессия.\]

\[\textbf{в)}\ x_{n} = n²\]

\[x_{1} = 1^{2} = 1\]

\[x_{2} = 2^{2} = 4\]

\[x_{3} = 3² = 9\]

\[q = \frac{x_{n}}{x_{n - 1}} = \frac{n^{2}}{(n - 1)^{2}} \Longrightarrow зависит\ \]

\[от\ n,\ не\ геометрическая\ \]

\[прогрессия.\]

\[\textbf{г)}\ x_{n} = ab^{n},\ \ где\ a \neq 0,\]

\[\ \ b \neq 0,\]

\[q = \frac{x_{n}}{x_{n - 1}} = \frac{ab^{n}}{ab^{n - 1}} = b^{n - n + 1} =\]

\[= b \Longrightarrow геометрическая\ \]

\[прогрессия.\]