Решебник по алгебре 9 класс Мерзляк Задание 917

\[\boxed{\mathbf{917\ (917).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[S = a^{2};\ \ где\ a - сторона\]

\[\ первого\ квадрата;\]

\[a_{1}^{2} = \left( \frac{a}{2} \right)^{2} + \left( \frac{a}{2} \right)^{2} = 2\frac{a^{2}}{4}\]

\[a_{2}^{2} = \frac{a^{2}}{2} \Longrightarrow \text{\ \ }a_{2} = \sqrt{\frac{a^{2}}{2}} =\]

\[= \frac{a}{\sqrt{2}} - второй\ квадрат;\]

\[S_{2} = a_{2}^{2} = \frac{a^{2}}{2};\ \]

\[a_{3}^{2} = \left( \frac{a}{2\sqrt{2}} \right)^{2} + \frac{a^{2}}{\left( 2\sqrt{2} \right)^{2}} =\]

\[= 2 \cdot \left( \frac{a}{2\sqrt{2}} \right)^{2}\text{\ \ }\]

\[a_{3}^{2} = 2\frac{a^{2}}{8}\text{\ \ }\]

\[a_{3}^{2} = \frac{a^{2}}{4} \Longrightarrow a_{3} = \frac{a}{2}\]

\[S_{3} = \frac{a^{2}}{4}\text{\ \ \ \ \ }и\ так\ далее.\]

\[Тогда\ \ a^{2},\frac{a^{2}}{2},\frac{a^{2}}{4},\ldots -\]

\[бесконечная\]

\[\ геометрическая\ прогрессия,\]

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

\[= \frac{a^{2}}{1 - \frac{1}{2}} = \frac{a^{2}}{\frac{1}{2}} = 2a^{2}.\]

\[Ответ:S = 2a².\]