Страница 235 номер 871, ГДЗ по алгебре за 9 класс к учебнику Мерзляка

Сумма \(n\) первых членов геометрической прогрессии

\(S_{n} = \dfrac{b_{1}(q^{n} - 1)}{q - 1}\)

1) \(b_{1} = 1,q = 2,n = 9\)
\(S_{9} = \dfrac{1 \cdot \left( 2^{9} - 1 \right)}{2 - 1} = 2^{9} - 1 = 512 - 1 = 511\)

2) \(b_{1} = 15,q = \frac{2}{3},n = 3\)
\(S_{3} = \dfrac{15 \cdot \left( \left( \frac{2}{3} \right)^{3} - 1 \right)}{\dfrac{2}{3} - 1} = \dfrac{15 \cdot \left( \frac{8}{27} - 1 \right)}{- \dfrac{1}{3}} =\)
\(= 15 \cdot \left( - \dfrac{19}{27} \right) \cdot ( - 3) = \dfrac{5 \cdot 19}{3} = \dfrac{95}{3}\)

3) \(b_{1} = 18,q = - \frac{1}{3},n = 5\)
\(S_{5} = \dfrac{18 \cdot \left( \left( - \frac{1}{3} \right)^{5} - 1 \right)}{- \dfrac{1}{3} - 1} = \dfrac{18 \cdot \left( - \frac{1}{3^{5}} - 1 \right)}{- \dfrac{4}{3}} =\)
\(= 18 \cdot \left( - \dfrac{1 + 3^{5}}{3^{5}} \right) \cdot \left( - \dfrac{3}{4} \right) = \dfrac{2 \cdot 9 \cdot \left( 1 + 3^{5} \right) \cdot 3}{3^{5} \cdot 4} = \dfrac{1 + 3^{5}}{3^{2} \cdot 2} = \dfrac{244}{9 \cdot 2} =\)
\(= \dfrac{122}{9}\)

4) \(b_{1} = 4,q = - \sqrt{2},n = 4\)
\(S_{4} = \dfrac{4 \cdot (\left( - \sqrt{2} \right)^{4} - 1)}{- \sqrt{2} - 1} = \dfrac{4 \cdot (4 - 1)}{- \sqrt{2} - 1} = - \dfrac{12}{\sqrt{2} + 1} =\)
\(= - \dfrac{12 \cdot \left( \sqrt{2} - 1 \right)}{2 - 1} = - 12(\sqrt{2} - 1)\)

Ответ: 1) \(S_{9} = 511\); 2) \(S_{3} = \frac{95}{3}\); 3) \(S_{5} = \frac{122}{9}\); 4) \(S_{4} = - 12(\sqrt{2} - 1)\).