Решебник по алгебре 9 класс Макарычев Задание 648

 

\[\boxed{\text{648}\text{\ (648)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ b_{1} = 8;\ \ q = \frac{1}{2}:\ \ \]

\[S_{5} = b_{1} \cdot \frac{q^{5} - 1}{q - 1} = 8 \cdot \frac{\left( \frac{1}{2} \right)^{5} - 1}{\frac{1}{2} - 1} = 8 \cdot \frac{\frac{1}{32} - 1\ }{- \frac{1}{2}} = \frac{8 \cdot 2 \cdot 31}{32} = 15,5.\]

\[\textbf{б)}\ b_{1} = 500;\ \ q = \frac{1}{5}:\ \ \]

\[S_{5} = b_{1} \cdot \frac{q^{5} - 1}{q - 1} = 500 \cdot \frac{\left( \frac{1}{5} \right)^{5} - 1}{\frac{1}{5} - 1\ } = \frac{500 \cdot 5 \cdot 3124}{4 \cdot 3125} = \frac{3124}{5} = 624,8.\]

\[\boxed{\text{648.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[Дано:\left( y_{n} \right) - арифметическая\ \]

\[прогрессия.\]

\[Доказать:\]

\[\textbf{а)}\ y_{2} + y_{7} = y_{4} + y_{5}\]

\[y_{2} = y_{1} + d,\]

\[y_{7} = y_{1} + 6d,\]

\[y_{4} = y_{1} + 3d\]

\[y_{5} = y_{1} + 4d\]

\[y_{1} + d + y_{1} + 6d =\]

\[= y_{1} + 3d + y_{1} + 4d\]

\[2y_{1} + 7d = 2y_{1} + 7d \Longrightarrow верно.\]

\[\textbf{б)}\ y_{n - 5} + y_{n + 10} = y_{n} + y_{n + 5},\]

\[\text{\ \ }где\ \ n > 5.\]

\[y_{n + 10} = y_{1} + d(n + 9)\]

\[y_{n - 5} = y_{1} + d(n - 6)\]

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

\[y_{n + 5} = y_{1} + d(n + 4)\]

\[y_{1} + d(n - 6) + y_{1} + d(n + 9) =\]

\[= y_{1} + d(n - 1) + y_{1} + d(n + 4)\]

\[2y_{1} + d(2n + 3) =\]

\[= 2y_{1} + d(2n + 3) \Longrightarrow ч.т.д.\]