ГДЗ по алгебре 9 класс Макарычев Задание 652

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

\[\textbf{а)}\ 1;3;3^{2};\ldots.\ \]

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

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

\[\textbf{б)}\ 2;2^{2};2^{3};\ldots\]

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

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

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

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

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

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

\[x \neq - 1\]

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

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

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

\[x \neq \pm 1\]

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

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

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

\[x \neq - 1\]

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

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