Решебник по алгебре 9 класс Мерзляк Задание 718

\[\boxed{\mathbf{718\ (718).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[a_{n} = a_{1} + d(n - 1)\]

\[a_{1} = 17;\ \ \ \ \ d = - 2\]

\[1)\ a_{4} = 17 - 2 \cdot 3 = 17 - 6 = 11\]

\[2)\ a_{15} = 17 - 2 \cdot 14 =\]

\[= 17 - 28 = - 11\]

\[3)\ a_{60} = 17 - 2 \cdot 59 =\]

\[= 17 - 118 = - 101\]

\[\boxed{\mathbf{718.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[b_{1} = 2\sqrt{3};\ \ q = \sqrt{3}\]

\[b_{1}^{2} + b_{2}^{2} + b_{3}^{2} + b_{4}^{2} + b_{5}^{2} + b_{6}^{2} =\]

\[= b_{1}^{2} + b_{1}^{2}q^{2} + b_{1}^{2q^{4}} + b_{1}^{2q^{6}} +\]

\[+ b_{1}^{2} \cdot q^{8} + b_{1}^{2} \cdot q^{10}\ \]

\[b_{1}^{2} = \left( 2\sqrt{3} \right)^{2} = 4 \cdot 3 = 12,\]

\[\ \ q = \frac{b_{1}^{2}q^{2}}{b_{1}²} = q² = \left( \sqrt{3} \right)^{2} = 3\]

\[S_{6} = \frac{12 \cdot \left( 3^{6} - 1 \right)}{3 - 1} =\]

\[= 6 \cdot (729 - 1) = 6 \cdot 728 =\]

\[= 4368.\]

\[Ответ:4368.\]