Страница 111 номер 306, ГДЗ по алгебре за 9 класс к учебнику Алимова с ответами и решением

1) \(b_{1} = 128\),
\(b_{2} = 64,\)
\(q = \dfrac{b_{2}}{b_{1}} = \dfrac{64}{128} = \dfrac{1}{2},\)
\(n = 6,\)
\(S_{n} = \dfrac{b_{1}\left(1 - q^{n}\right)}{1 - q} = \dfrac{128\left(1 - \left(\frac{1}{2}\right)^{6}\right)}{1 - \dfrac{1}{2}} = 256 \cdot \dfrac{63}{64} = 252;\)

2) \(b_{1} = 162\),
\(b_{2} = 54,\)
\(q = \dfrac{b_{2}}{b_{1}} = \dfrac{54}{162} = \dfrac{1}{3},\)
\(n = 5,\)
\(S_{n} = \dfrac{b_{1}\left(1 - q^{n}\right)}{1 - q} = \dfrac{162\left(1 - \left(\frac{1}{3}\right)^{5}\right)}{1 - \dfrac{1}{3}} = 243 \cdot \dfrac{242}{243} = 242;\)

3) \(b_{1} = \frac{2}{3}\),
\(b_{2} = \dfrac{1}{2},\)
\(q = \dfrac{b_{2}}{b_{1}} = \dfrac{1}{2} \cdot \dfrac{3}{2} = \dfrac{3}{4},\)
\(n = 5,\)
\(S_{n} = \dfrac{b_{1}\left(1 - q^{n}\right)}{1 - q} = \dfrac{\frac{2}{3} \cdot \left(1 - \left(\frac{3}{4}\right)^{5}\right)}{1 - \dfrac{3}{4}} = \dfrac{8}{3} \cdot \dfrac{781}{1024} = \dfrac{781}{384} = 2\dfrac{13}{384};\)

4) \(b_{1} = \frac{3}{4}\),
\(b_{2} = \dfrac{1}{2},\)
\(q = \dfrac{b_{2}}{b_{1}} = \dfrac{1}{2} \cdot \dfrac{4}{3} = \dfrac{2}{3},\)
\(n = 4,\)
\(S_{n} = \dfrac{b_{1}\left(1 - q^{n}\right)}{1 - q} = \dfrac{\frac{3}{4}\left(1 - \left(\frac{2}{3}\right)^{4}\right)}{1 - \dfrac{2}{3}} = \dfrac{9}{4} \cdot \dfrac{65}{81} = \dfrac{65}{36} = 1\dfrac{29}{36};\)

5) \(b_{1} = -\frac{1}{8}\),
\(b_{2} = -\dfrac{1}{2},\)
\(q = \dfrac{b_{2}}{b_{1}} = \dfrac{1}{2} \cdot 8 = 4,\)
\(n = 4,\)
\(S_{n} = \dfrac{b_{1}\left(1 - q^{n}\right)}{1 - q} = \dfrac{-\frac{1}{8}\left(1 - 4^{4}\right)}{1 - 4} =\)
\(= \dfrac{1}{24} \cdot (1 - 256) = -\dfrac{255}{24} = -10\dfrac{5}{8};\)

6) \(b_{1} = -4\),
\(b_{2} = 8,\)
\(q = \dfrac{b_{2}}{b_{1}} = -\dfrac{8}{4} = -2,\)
\(n = 5,\)
\(S_{n} = \dfrac{b_{1}\left(1 - q^{n}\right)}{1 - q} = \dfrac{-4\left(1 - (-2)^{5}\right)}{1 - (-2)} = -\dfrac{4}{3} \cdot 33 = -44.\)