Страница 235 номер 870, 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) \(b_{1} = 10,q = 3,n = 4\)
\(S_{4} = \dfrac{10 \cdot (3^{4} - 1)}{3 - 1} = 5 \cdot (81 - 1) = 5 \cdot 80 = 400\)

2) \(b_{1} = - 4,q = - 1,n = 10\)
\(S_{10} = \dfrac{- 4 \cdot (( - 1)^{10} - 1)}{- 1 - 1} = \dfrac{- 4 \cdot (1 - 1)}{- 2} = 0\)

3) \(b_{1} = 0{,}6,\ q = 2,\ n = 5\)
\(S_{5} = \dfrac{0{,}6 \cdot (2^{5} - 1)}{2 - 1} = 0{,}6 \cdot 31 = 18{,}6\)

4) \(b_{1} = 4{,}5,\ q = \frac{1}{3},n = 8\)
\(S_{8} = \dfrac{4{,}5\left( \frac{1}{3^{8}} - 1 \right)}{\dfrac{1}{3} - 1\ } = 4{,}5\ \left( \dfrac{1 - 3^{8}}{3^{8}} \right) \cdot \left( - \dfrac{3}{2} \right) =\)
\(= \dfrac{45}{10} \cdot \dfrac{- 6560}{3^{8}} \cdot \left( - \dfrac{3}{2} \right) = 45 \cdot \dfrac{328}{3^{7}} = \dfrac{5 \cdot 3^{2} \cdot 328}{3^{7}} = \dfrac{1640}{243}\)

5) \(b_{1} = - 9,q = \sqrt{3},\ n = 6\)
\(S_{6} = \dfrac{- 9({\sqrt{3}}^{6} - 1)}{\sqrt{3} - 1} = \dfrac{- 9\left( 3^{3} - 1 \right)}{\sqrt{3} - 1} = \dfrac{- 9 \cdot 26}{\sqrt{3} - 1} =\)
\(= \dfrac{- 9 \cdot 26 \cdot \left( \sqrt{3} + 1 \right)}{3 - 1} = - 9 \cdot 13 \cdot \left( \sqrt{3} + 1 \right) = - 117(\sqrt{3} + 1)\)

6) \(b_{1} = 8,q = - \frac{1}{2},n = 4\)
\(S_{4} = \dfrac{8 \cdot \left( \left( - \frac{1}{2} \right)^{4} - 1 \right)}{- \dfrac{1}{2} - 1} = 8 \cdot \left( \dfrac{1}{16} - 1 \right) \cdot \left( - \dfrac{2}{3} \right) =\)
\(= 8 \cdot \left( - \dfrac{15}{16} \right) \cdot \left( - \dfrac{2}{3} \right) = 5\)

Ответ: 1) \(S_{4} = 400\); 2) \(S_{10} = 0\); 3) \(S_{5} = 18{,}6\); 4) \(S_{8} = \frac{1640}{243}\); 5) \(S_{6} = - 117(\sqrt{3} + 1)\); 6) \(S_{4} = 5\).