ГДЗ по алгебре 9 класс Макарычев Задание 597

 

\[\boxed{\text{597\ (}\text{с}\text{).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[a_{n} = kn + b;\ \ \]

\[где\ \text{k\ }и\ b - некоторые\ числа,\ \]

\[является\]

\[арифметической\ прогрессией.\]

\[\textbf{а)}\ a_{n} = 3n + 1:\ \]

\[k = 3;\ \ b = 1;\ \]

\[d = k = 3\ \ \]

\[a_{1} = 3 + 1 = 4.\]

\[\textbf{б)}\ a_{n} = n² - 5,\ \ \]

\[a_{n} - a_{n - 1} =\]

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

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

\[= 2n + 1\ \ \]

\[зависит\ от\ n \Longrightarrow не\ является\ \]

\[арифметической\ прогрессией.\]

\[\textbf{в)}\ a_{n} = n + 4:\ \]

\[k = d = 1\ \ \]

\[a_{1} = 1 + 4 = 5.\]

\[\textbf{г)}\ a_{n} = \frac{1}{n + 4}:\ \ \]

\[a_{n} - a_{n - 1} = \frac{1}{n + 4} - \frac{1}{n + 3} =\]

\[= \frac{n + 3 - n - 4}{(n + 4)(n + 3)} =\]

\[= \frac{- 1}{(n + 4)(n + 3)} \Longrightarrow\]

\[\Longrightarrow зависит\ от\ n \Longrightarrow не\ \]

\[прогрессия;\]

\[\textbf{д)}\ a_{n} = - 0,5n + 1:\ \ \]

\[k = d = - 0,5;\ \ \ \]

\[a_{1} = - 0,5 + 1 = 0,5.\]

\[\textbf{е)}\ a_{n} = 6:\ \ \]

\[a_{n} - a_{n - 1} = 6n - 6 \cdot (n - 1) =\]

\[= 6;\]

\[d = k = 6;\ \ \]

\[a_{1} = 6.\]

\[\boxed{\text{597.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ c_{5} = c_{1} \cdot q^{4} = - 6,\ \ \]

\[c_{7} = c_{1} \cdot q^{6} = - 54,\ \]

\[\Longrightarrow \frac{c_{1} \cdot q^{6}}{c_{1} \cdot q^{4}} = \frac{- 54}{- 6};\ \ \ \ q^{2} = 9,\]

\[\ \ q = \pm 3.\]

\[\textbf{б)}\ c_{6} = c_{1} \cdot q^{5} = 25,\ \ \]

\[c_{8} = c_{1} \cdot q^{7} = 4,\ \]

\[\frac{c_{8} \cdot q^{7}}{c_{6} \cdot q^{5}} = \frac{4}{25},\ \ q^{2} = \frac{4}{25},\]

\[\ \ q = \pm \frac{2}{5}.\]