17.28. Найдите сумму первых пяти членов геометрической прогрессии:

Привет! Давай разберем задачи на геометрическую прогрессию. Формула суммы первых n членов: $S_n = \frac{b_1(q^n - 1)}{q - 1}$. ### Решение 17.28 а) $b_1 = 3, q = 2$. $S_5 = \frac{3(2^5 - 1)}{2 - 1} = 3(32 - 1) = 3 \cdot 31 = 93$. б) $b_1 = -1, q = -2$. $S_5 = \frac{-1((-2)^5 - 1)}{-2 - 1} = \frac{-1(-32 - 1)}{-3} = \frac{33}{-3} = -11$. в) $b_1 = -3, q = 1/2$. $S_5 = \frac{-3((1/2)^5 - 1)}{1/2 - 1} = \frac{-3(1/32 - 1)}{-1/2} = \frac{-3(-31/32)}{-1/2} = -3 \cdot \frac{31}{32} \cdot (-2) = \frac{3 \cdot 31}{16} = \frac{93}{16} = 5,8125$. г) $b_1 = \sqrt{2}, q = 3$. $S_5 = \frac{\sqrt{2}(3^5 - 1)}{3 - 1} = \frac{\sqrt{2}(243 - 1)}{2} = \frac{\sqrt{2} \cdot 242}{2} = 121\sqrt{2}$. ### Решение 17.29 а) $b_4=160, b_5=320 \Rightarrow q = \frac{320}{160} = 2$. $b_1 = \frac{b_4}{q^3} = \frac{160}{8} = 20$. $S_5 = \frac{20(2^5 - 1)}{2 - 1} = 20 \cdot 31 = 620$. б) $b_7=8, b_9=16, q<0$. $b_9 = b_7 \cdot q^2 \Rightarrow 16 = 8 \cdot q^2 \Rightarrow q^2 = 2 \Rightarrow q = -\sqrt{2}$. $b_1 = \frac{b_7}{q^6} = \frac{8}{(-\sqrt{2})^6} = \frac{8}{8} = 1$. $S_5 = \frac{1((-\sqrt{2})^5 - 1)}{-\sqrt{2} - 1} = \frac{-4\sqrt{2} - 1}{-\sqrt{2} - 1} = \frac{4\sqrt{2} + 1}{\sqrt{2} + 1} = \frac{(4\sqrt{2}+1)(\sqrt{2}-1)}{2-1} = 8 - 4\sqrt{2} + \sqrt{2} - 1 = 7 - 3\sqrt{2}$. в) $b_3=1, b_5=1/9, q>0$. $b_5 = b_3 \cdot q^2 \Rightarrow 1/9 = 1 \cdot q^2 \Rightarrow q = 1/3$. $b_1 = \frac{b_3}{q^2} = \frac{1}{1/9} = 9$. $S_5 = \frac{9((1/3)^5 - 1)}{1/3 - 1} = \frac{9(1/243 - 1)}{-2/3} = 9 \cdot (-\frac{242}{243}) \cdot (-\frac{3}{2}) = \frac{242}{27} \cdot \frac{3}{2} = \frac{121}{9} = 13\frac{4}{9}$. г) $b_4=3\sqrt{3}, b_7=27$. $b_7 = b_4 \cdot q^3 \Rightarrow 27 = 3\sqrt{3} \cdot q^3 \Rightarrow q^3 = \frac{27}{3\sqrt{3}} = \frac{9}{\sqrt{3}} = 3\sqrt{3} = (\sqrt{3})^3 \Rightarrow q = \sqrt{3}$. $b_1 = \frac{b_4}{q^3} = \frac{3\sqrt{3}}{3\sqrt{3}} = 1$. $S_5 = \frac{1((\sqrt{3})^5 - 1)}{\sqrt{3} - 1} = \frac{9\sqrt{3} - 1}{\sqrt{3} - 1} = \frac{(9\sqrt{3}-1)(\sqrt{3}+1)}{3-1} = \frac{27 + 9\sqrt{3} - \sqrt{3} - 1}{2} = \frac{26 + 8\sqrt{3}}{2} = 13 + 4\sqrt{3}$.