Решебник по алгебре 9 класс Мерзляк Задание 704

 

\[\boxed{\mathbf{704\ (704).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[a_{n} = n^{2} - 3n - 8\ \ \]

\[n^{2} - 3n - 8 < 10\ \ \]

\[n^{2} - 3n - 18 < 0\]

\[n_{1} + n_{2} = 3,\ \ n_{1} = 6\]

\[n_{1}n_{2} = - 18,\ \ n_{2} = - 3\]

\[Так\ как\ номера - это\ \]

\[положительные\ числа,\ \]

\[тогда\ n \in \lbrack 1;6).\]

\[Ответ:n = \{ 1;2;3;4;5\}.\]

\[\boxed{\mathbf{704.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ b_{1} = 1,\ \ q = 2,\ \ n = 9:\ \ \]

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

\[= 512 - 1 = 511;\]

\[2)\ b_{1} = 15,\ \ q = \frac{2}{3},\]

\[\ \ n = 3:\ \]

\[S_{3} = \frac{15 \cdot \left( \left( \frac{2}{3} \right)^{3} - 1 \right)}{\frac{2}{3} - 1\ } =\]

\[\frac{15 \cdot \left( \frac{8}{27} - 1 \right)}{- \frac{1}{3}} = \frac{15 \cdot 19 \cdot 3}{27 \cdot 1} = \frac{95}{3};\]

\[3)\ b_{1} = 18,\ \ q = - \frac{1}{3},\]

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

\[S_{5} = \frac{18 \cdot \left( \left( - \frac{1}{3} \right)^{5} - 1 \right)}{- \frac{1}{3} - 1} =\]

\[= \frac{18 \cdot \left( - \frac{1}{243} - 1 \right)\ }{- \frac{4}{3}} =\]

\[= \frac{18 \cdot 244 \cdot 3}{243 \cdot 4} = \frac{122}{9};\]

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

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

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

\[= \frac{4 \cdot (4 - 1)}{- \sqrt{2} - 1} = \frac{4 \cdot 3}{- \sqrt{2} - 1} =\]

\[= \frac{12 \cdot \left( \sqrt{2} - 1 \right)}{- \left( \sqrt{2} + 1 \right)\left( \sqrt{2} - 1 \right)} =\]

\[= \frac{12 \cdot \left( \sqrt{2} - 1 \right)}{- (2 - 1)} =\]

\[= \frac{12 \cdot \left( \sqrt{2} - 1 \right)}{- 1} =\]

\[= - 12 \cdot \left( \sqrt{2} - 1 \right) =\]

\[= 12 \cdot \left( 1 - \sqrt{2} \right).\]