Решебник по алгебре 9 класс Макарычев Задание 651

 

\[\boxed{\text{651\ (}\text{c}\text{).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ b_{n} = 0,2 \cdot 5^{n},\ \ q = \frac{b_{n}}{b_{n - 1}} = \frac{0,2 \cdot 5^{n}}{0,2 \cdot 5^{n - 1}} = 5,\ \ b_{1} = 0,2 \cdot 5 = 1,\]

\[\Longrightarrow b_{n} - геометрическая\ прогрессия;\]

\[S_{n} = b_{1} \cdot \frac{q^{n} - 1}{q - 1} = \frac{5^{n} - 1}{5 - 1} = \frac{5^{n} - 1}{4};\]

\[\textbf{б)}\ b_{n} = 3 \cdot 2^{n - 1},\ \ q = \frac{b_{n}}{b_{n - 1}} = \frac{3 \cdot 2^{n - 1}}{3 \cdot 2^{n - 2}} = 2,\ \ b_{1} = 3\]

\[\Longrightarrow b_{n} - геометрическая\ прогрессия;\]

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

\[\textbf{в)}\ b_{n} = 3^{1 + n},\ \ q = \frac{b_{n}}{b_{n - 1}} = \frac{3^{1 + n}}{3^{n}} = 3,\ \ b_{1} = 9\]

\[\Longrightarrow b_{n} - геометрическая\ прогрессия;\]

\[S_{n} = b_{1} \cdot \frac{q^{n} - 1}{q - 1} = 9 \cdot \frac{3^{n} - 1}{3 - 1} = \frac{9}{2} \cdot (3^{n} - 1).\]

\[\boxed{\text{651.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ a_{1} = \frac{2}{3};\ \ \ a_{2} = \frac{3}{4} \Longrightarrow d =\]

\[= a_{2} - a_{1} = \frac{3}{4} - \frac{2}{3} = \frac{1}{12};\]

\[S_{10} = \frac{2a_{1} + d(n - 1)}{2} \cdot n =\]

\[= \frac{2 \cdot \frac{2}{3} + \frac{1}{12} \cdot (10 - 1)}{2} \cdot 10 =\]

\[= \left( \frac{4}{3} + \frac{3}{4} \right) \cdot 5 =\]

\[= \frac{16 + 9}{12} \cdot 5 = \frac{25}{12} \cdot 5 = \frac{125}{12} =\]

\[= 10\frac{5}{12}.\]

\[\textbf{б)}\ a_{1} = \sqrt{3},\ \ a_{2} = \sqrt{12},\ \]

\[\Longrightarrow d = a_{2} - a_{1} = \sqrt{12} - \sqrt{3} =\]

\[= \sqrt{3} \cdot (2 - 1) = \sqrt{3};\]

\[S_{10} = \frac{2a_{1} + d(n - 1)}{2} \cdot n =\]

\[= \left( 2\sqrt{3} + 9\sqrt{3} \right) \cdot 5 = 11\sqrt{3} \cdot 5 =\]

\[= 55\sqrt{3}.\]