ГДЗ по алгебре 9 класс Макарычев Задание 652

 

\[\boxed{\text{652\ (652).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ 1;3;3^{2};\ldots.\ \]

\[b_{1} = 1;\ \ q = 3,\ \ \]

\[S_{n} = b_{1} \cdot \frac{q^{n} - 1}{q - 1} = \frac{3^{n} - 1}{3 - 1} = \frac{3^{n} - 1}{2}.\]

\[\textbf{б)}\ 2;2^{2};2^{3};\ldots\]

\[b_{1} = 2,\ \ q = 2,\ \ \]

\[S_{n} = b_{1} \cdot \frac{q^{n} - 1}{q - 1} = 2 \cdot \frac{2^{n} - 1}{2 - 1} = 2^{n + 1} - 2.\]

\[\textbf{в)}\ \frac{1}{2}; - \frac{1}{4};\ \ \frac{1}{8};\ldots.\]

\[b_{1} = \frac{1}{2},\ \ q = - \frac{1}{2},\ \ \]

\[S_{n} = b_{1} \cdot \frac{q^{n} - 1}{q - 1} = \frac{1}{2} \cdot \frac{\left( - \frac{1}{2} \right)^{n} - 1}{- \frac{1}{2} - 1} = - \frac{\left( - \frac{1}{2} \right)^{n} - 1}{3} = - \frac{1}{3} \cdot \left( \left( - \frac{1}{2} \right)^{n} - 1 \right).\]

\[\textbf{г)}\ 1;\ - x;\ \ x^{2};\ \ldots.\]

\[x \neq - 1\]

\[b_{1} = 1,\ \ q = - x,\ \ \]

\[S_{n} = \frac{( - x)^{n} - 1}{- x - 1} = \frac{1 - ( - x)^{n}}{x + 1}.\]

\[\textbf{д)}\ 1;\ \ x^{2};\ \ x^{4};\ldots.\ \ \]

\[x \neq \pm 1\]

\[b_{1} = 1,\ \ q = x^{2},\ \ b_{2} = x^{2};\ \ \]

\[S_{n} = \frac{x^{2n} - 1}{x² - 1}.\]

\[\textbf{е)}\ 1; - x^{3};\ \ x^{6};\ldots.\]

\[x \neq - 1\]

\[b_{1} = 1,\ \ q = - x^{3},\ \ \]

\[S_{n} = \frac{( - x)^{3n} - 1}{- x^{3} - 1} = \frac{1 - ( - x)^{3n}}{x^{3} + 1}\text{.\ }\]

\[\boxed{\text{652.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ a_{1} = 2,\ \ a_{2} = 6,\]

\[\ \ d = a_{2} - a_{1} = 4,\]

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

\[198 = 2 + 4n - 4\]

\[4n = 200\]

\[n = 50;\]

\[S_{50} = \frac{a_{1} + a_{50}}{2} \cdot n =\]

\[= \frac{2 + 198}{2} \cdot 50 = 100 \cdot 50 =\]

\[= 5000.\]

\[\textbf{б)}\ a_{1} = 95,\ \ a_{2} = 85,\]

\[\ \ d = a_{2} - a_{1} = - 10,\ \]

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

\[- 155 = 95 - 10n + 10\]

\[10n = 260\]

\[n = 26;\]

\[S_{26} = \frac{a_{1} + a_{26}}{2} \cdot n =\]

\[= \frac{95 - 155}{2} \cdot 26 = - 60 \cdot 13 =\]

\[= - 780.\]