Решебник по алгебре 9 класс Макарычев Задание 659

\[\boxed{\text{659}\text{\ (659)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \frac{2^{n + 2} - 2^{n - 1}}{2^{n}} = \frac{2^{n} \cdot \left( 2^{2} - 2^{- 2} \right)}{2^{n}} = 2² - \frac{1}{4} = 4 - \frac{1}{4} = 3\frac{3}{4};\]

\[\textbf{б)}\ \frac{25^{n} - 5^{2n - 1}}{5^{2n}} = \frac{5^{2n} - 5^{2n - 1}}{5^{2n}} = \frac{5^{2n} \cdot (1 - 5^{- 1})}{5^{2n}} = 1 - \frac{1}{5} = \frac{4}{5}.\]

\[\boxed{\text{659.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ a_{1} = 1,\ \ d = 1,\]

\[a_{n} = a_{1} + d(n - 1) \Longrightarrow\]

\[\Longrightarrow 1 + n - 1 = n \Longrightarrow\]

\[S_{n} = \frac{a_{1} + a_{n}}{2} \cdot n = \frac{1 + n}{2} \cdot n;\]

\[5a_{n + 1} = S_{n}\]

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

\[n = 10\]

\[a_{n + 1} = n + 1 = 11.\]

\[\textbf{б)}\ a_{1} = 1,\ \ d = 1,\]

\[a_{n} = a_{1} + d(n - 1) =\]

\[= 1 + n - 1 = n,\]

\[S_{n} = \frac{a_{1} + a_{n}}{2} \cdot n = \frac{1 + n}{2} \cdot n\]

\[a_{n + 1} = S_{n}\]

\[n + 1 = \frac{1 + n}{2} \cdot n\]

\[n = 2 \Longrightarrow x_{n + 1} = n + 1 = 3.\]