ГДЗ по алгебре 9 класс Мерзляк Задание 870

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\[1)\ b_{1} = 10,\ \ q = 3,\]

\[\ \ n = 4:\ \ \]

\[S_{4} = \frac{10 \cdot \left( 3^{4} - 1 \right)}{3 - 1} =\]

\[= \frac{10 \cdot (81 - 1)}{2} = 5 \cdot 80 = 400;\]

\[2)\ b_{1} = - 4,\ \ q = - 1,\ \]

\[\ n = 10:\ \]

\[S_{10} = \frac{- 4 \cdot \left( ( - 1)^{9} - 1 \right)}{- 1 - 1} =\]

\[= \frac{- 4 \cdot ( - 1 - 1)}{- 2} = - 4;\]

\[3)\ b_{1} = 0,6,\ \ q = 2,\]

\[\ \ n = 5:\ \ \]

\[S_{5} = \frac{0,6 \cdot \left( 2^{5} - 1 \right)}{2 - 1} =\]

\[= \frac{0,6 \cdot (32 - 1)}{1} = 0,6 \cdot 31 =\]

\[= 18,6;\]

\[4)\ b_{1} = 4,5,\ \ q = \frac{1}{3},\]

\[\ \ n = 8:\ \ \]

\[S_{8} = \frac{4,5 \cdot \left( \left( \frac{1}{3} \right)^{8} - 1 \right)}{\frac{1}{3} - 1\ } =\]

\[= \frac{\frac{9}{2} \cdot \left( \frac{1}{6561} - 1 \right)}{- \frac{2}{3}} =\]

\[= \frac{9 \cdot 3 \cdot 6560}{2 \cdot 2 \cdot 6561} = \frac{1640}{243};\]

\[5)\ b_{1} = - 9,\ \ q = \sqrt{3},\]

\[\ \ n = 6:\ \ \]

\[S_{6} = \frac{- 9 \cdot \left( \left( \sqrt{3} \right)^{6} - 1 \right)}{\sqrt{3} - 1} =\]

\[= \frac{- 9 \cdot (27 - 1)}{\sqrt{3} - 1} =\]

\[= \frac{- 9 \cdot 26 \cdot \left( \sqrt{3} + 1 \right)}{\left( \sqrt{3} - 1 \right)\left( \sqrt{3} + 1 \right)} =\]

\[= \frac{- 9 \cdot 26 \cdot \left( \sqrt{3} + 1 \right)}{3 - 1} =\]

\[= - 9 \cdot 13 \cdot \left( \sqrt{3} + 1 \right) =\]

\[= - 117 \cdot (\sqrt{3} + 1)\]

\[6)\ b_{1} = 8,\ \ q = - \frac{1}{2},\]

\[\ \ n = 4:\ \ \]

\[S_{4} = \frac{8 \cdot \left( \left( - \frac{1}{2} \right)^{4} - 1 \right)}{- \frac{1}{2} - 1\ } =\]

\[= \frac{8 \cdot \left( \frac{1}{16} - 1 \right)}{- \frac{3}{2}} = \frac{8 \cdot 15 \cdot 2}{16 \cdot 3} = 5\]