Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 426

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\[1)\ q = \frac{1}{2}\text{\ \ }и\ \ b_{5} = \frac{\sqrt{2}}{16};\]

\[b_{5} = b_{1} \bullet q^{4}\]

\[b_{1} = \frac{b_{5}}{q^{4}};\]

\[b_{1} = \frac{\sqrt{2}}{16}\ :\left( \frac{1}{2} \right)^{4} = \frac{\sqrt{2}}{16}\ :\frac{1}{16} =\]

\[= \frac{\sqrt{2}}{16} \bullet 16 = \sqrt{2};\]

\[S = \frac{b_{1}}{1 - q} = \frac{\sqrt{2}}{1 - \frac{1}{2}} = \sqrt{2}\ :\frac{1}{2} =\]

\[= \sqrt{2} \bullet 2 = 2\sqrt{2}.\]

\[2)\ q = \frac{\sqrt{3}}{2}\text{\ \ }и\ \ b_{4} = \frac{9}{8};\]

\[b_{4} = b_{1} \bullet q^{3},\ отсюда\ b_{1} = \frac{b_{4}}{q^{3}};\]

\[b_{1} = \frac{9}{8}\ :\left( \frac{\sqrt{3}}{2} \right)^{3} = \frac{9}{8}\ :\frac{3\sqrt{3}}{8} =\]

\[= \frac{9}{8} \bullet \frac{8}{3\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3};\]

\[S = \frac{b_{1}}{1 - q} = \frac{\sqrt{3}}{1 - \frac{\sqrt{3}}{2}} =\]

\[= \sqrt{3}\ :\frac{2 - \sqrt{3}}{2} = \sqrt{3} \bullet \frac{2}{2 - \sqrt{3}};\]

\[S = \sqrt{3} \bullet \frac{2\left( 2 + \sqrt{3} \right)}{\left( 2 - \sqrt{3} \right)\left( 2 + \sqrt{3} \right)} =\]

\[= 2\sqrt{3} \bullet \frac{2 + \sqrt{3}}{4 - 3} =\]

\[= 2\sqrt{3}\left( 2 + \sqrt{3} \right) = 4\sqrt{3} + 6.\]