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МатематикаРешебник по алгебре 9 класс Мерзляк Задание 696
Задание 696
\[\boxed{\mathbf{696\ (696).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[c_{n} = ( - 1)^{n} \cdot 5\]
\[1)\ c_{1} = - 5\]
\[2)\ c_{8} = 5\]
\[3)\ c_{2k} = ( - 1)^{2k} \cdot 5 = 5 \cdot 1^{k} = 5\]
\[4)\ \ c_{2k + 1} = ( - 1)^{2k + 1} \cdot 5 =\]
\[= ( - 1)^{2k} \cdot ( - 1)^{1} \cdot 5 = - 5\]
\[5)\ \ c_{k + 2} = ( - 1)^{k + 2} \cdot 5 =\]
\[= ( - 1)^{k} \cdot ( - 1)^{2} \cdot 5 = ( - 1)^{k} \cdot 5\]
\[\boxed{\mathbf{696.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[Геометрическая\ прогрессия:\]
\[b_{1},\ b_{2},\ b_{3},\ \ b_{1} + b_{2} + \ b_{3} = 65,\]
\[b_{1} + b_{1}q + b_{1}q^{2} = 65,\ \ \]
\[b_{1}\left( 1 + q + q^{2} \right) = 65.\]
\[Арифметическая\ прогрессия:\ \]
\[b_{1} - 1,b_{2},\ b_{3} - 19\ \]
\[b_{2} = \frac{b_{1} - 1 + b_{3} - 19}{2}\]
\[\ 2b_{2} = b_{1} + b_{3} - 20\]
\[\ 2b_{2} - b_{1} - b_{3} = - 20\]
\[b_{1} + b_{3} - 2b_{2} = 20\]
\[b_{1} + b_{1}q^{2} - 2b_{1}q = 20\ \ \]
\[b_{1}\left( 1 + q^{2} - 2q \right) = 20.\]
\[Запишем\ систему:\]
\[\left\{ \begin{matrix} b_{1}\left( 1 + q^{2} - 2q \right) = 20 \\ b_{1}\left( 1 + q + q^{2} \right) = 65 \\ \end{matrix}\ |\ (\ :)\text{\ \ \ \ } \right.\ \]
\[\frac{b_{1}\left( 1 + q^{2} - 2q \right)}{b_{1}\left( 1 + q + q^{2} \right)} = \frac{20}{65}\]
\[\ \frac{1 + q^{2} - 2q}{1 + q + q^{2}} = \frac{4}{13}\]
\[13 \cdot \left( 1 + q^{2} - 2q \right) =\]
\[= 4 \cdot \left( 1 + q + q^{2} \right)\]
\[13 + 13q^{2} - 23q =\]
\[= 4 + 4q + 4q^{2}\]
\[9q² - 30q + 9 = 0\ \ \ \ |\ :3\]
\[3q^{2} - 10q + 3 = 0\]
\[D = 100 - 36 = 64\]
\[q_{1} = \frac{10 + 8}{6} = 3;\ \]
\[\ q_{2} = \frac{10 - 8}{6} = \frac{1}{3}\]
\[b_{1} \cdot \left( 1 + q^{2} - 2q \right) = 20\ \]
\[b_{1}(1 - q)^{2} = 20\ \ \]
\[b_{1} = \frac{20}{(1 - q)^{2}}\]
\[при\ \ q = 3:\ \ \]
\[b_{1} = \frac{20}{(1 - 3)^{2}} = \frac{20}{4} = 5,\]
\[\text{\ \ }b_{2} = 5 \cdot 3 = 15,\ \ \]
\[b_{3} = 15 \cdot 3 = 45;\]
\[при\ q = \frac{1}{3}:\ \ \]
\[b_{1} = \frac{20}{\left( 1 - \frac{1}{3} \right)^{2}} = \frac{20 \cdot 9}{4} = 45,\]
\[\text{\ \ }b_{2} = 45 \cdot \frac{1}{3} = 15,\]
\[\text{\ \ }b_{3} = 15 \cdot \frac{1}{3} = 5\]
\[Ответ:5,\ 15,\ 45\ \ \ или\ \ \ 45,\ 15,\ 5.\]
