Страница 243 номер 902, ГДЗ по алгебре за 9 класс к учебнику Мерзляка
Найдите сумму бесконечной геометрической прогрессии: 1) \(\sqrt{2},\ -1,\ \frac{1}{\sqrt{2}},\ \ldots\); 2) \(3\sqrt{3},\ 3,\ \sqrt{3},\ \ldots\); 3) \(\frac{\sqrt{3} + 1}{\sqrt{3} - 1},\ 1,\ \frac{\sqrt{3} - 1}{\sqrt{3} + 1},\ \ldots\) .
1) \(\sqrt{2}, - 1,\frac{1}{\sqrt{2}},\ldots\)
Первый член прогрессии \(b_{1} = \sqrt{2}\)
Знаменатель прогрессии \(q = - \frac{1}{\sqrt{2}}\)
\(S = \dfrac{b_{1}}{1 - q} = \dfrac{\sqrt{2}}{1 - \left( - \frac{1}{\sqrt{2}} \right)} = \dfrac{\sqrt{2}}{\frac{\sqrt{2} + 1}{\sqrt{2}}} =\)
\(= \sqrt{2} \cdot \dfrac{\sqrt{2}}{\sqrt{2} + 1} = \dfrac{2(\sqrt{2} - 1)}{2 - 1} = 2\left( \sqrt{2} - 1 \right)\)
2) \(3\sqrt{3},\ 3,\ \sqrt{3},\ \ldots\)
\(b_{1} = 3\sqrt{3}\)
\(q = \dfrac{3}{3\sqrt{3}} = \dfrac{1}{\sqrt{3}}\)
\(S = \dfrac{b_{1}}{1 - q} = \dfrac{3\sqrt{3}}{1 - \dfrac{1}{\sqrt{3}}} = 3\sqrt{3} \cdot \dfrac{\sqrt{3}}{\sqrt{3} - 1} = \dfrac{9\left( \sqrt{3} + 1 \right)}{3 - 1} = \dfrac{9}{2}\left( \sqrt{3} + 1 \right)\)
3) \(\frac{\sqrt{3} + 1}{\sqrt{3} - 1},\ 1,\frac{\sqrt{3} - 1}{\sqrt{3} + 1},\ \ldots\)
\(b_{1} = \dfrac{\sqrt{3} + 1}{\sqrt{3} - 1}\)
\(q = 1:\dfrac{\sqrt{3} + 1}{\sqrt{3} - 1} = \dfrac{\sqrt{3} - 1}{\sqrt{3} + 1}\)
\(S = \dfrac{b_{1}}{1 - q} = \dfrac{\sqrt{3} + 1}{\sqrt{3} - 1}:\left( 1 - \dfrac{\sqrt{3} - 1}{\sqrt{3} + 1} \right) =\)
\(= \dfrac{\sqrt{3} + 1}{\sqrt{3} - 1} \cdot \dfrac{\sqrt{3} + 1}{\sqrt{3} + 1 - \sqrt{3} + 1} = \dfrac{\left( \sqrt{3} + 1 \right)^{2}\left( \sqrt{3} + 1 \right)}{(3 - 1) \cdot 2} =\)
\(= \dfrac{\left( \sqrt{3} + 1 \right)^{3}}{4} = \dfrac{3\sqrt{3} + 3 \cdot 3 + 3\sqrt{3} + 1}{4} = \dfrac{6\sqrt{3} + 10}{4} = \dfrac{3\sqrt{3} + 5}{2}\)
Ответ: 1) \(S = 2\left( \sqrt{2} - 1 \right)\); 2) \(S = \frac{9}{2}\left( \sqrt{3} + 1 \right)\); 3) \(S = \frac{3\sqrt{3} + 5}{2}\).
