Найдите сумму n первых членов геометрической прогрессии (bn) со знаменателем q, если:

Для решения задач на сумму $n$ первых членов геометрической прогрессии $S_n$ используется формула: $S_n = \frac{b_1(q^n - 1)}{q - 1}$, где $q \neq 1$. Если $q = 1$, то $S_n = n \cdot b_1$. ### Решение упражнения 870 1) $b_1 = 10, q = 3, n = 4$ $S_4 = \frac{10(3^4 - 1)}{3 - 1} = \frac{10(81 - 1)}{2} = \frac{10 \cdot 80}{2} = 400$. 2) $b_1 = -4, q = -1, n = 10$ $S_{10} = \frac{-4((-1)^{10} - 1)}{-1 - 1} = \frac{-4(1 - 1)}{-2} = 0$. 3) $b_1 = 0.6, q = 2, n = 5$ $S_5 = \frac{0.6(2^5 - 1)}{2 - 1} = 0.6(32 - 1) = 0.6 \cdot 31 = 18.6$. 4) $b_1 = 4.5, q = 1/3, n = 8$ $S_8 = \frac{4.5((1/3)^8 - 1)}{1/3 - 1} = \frac{4.5(1/6561 - 1)}{-2/3} = \frac{4.5(-6560/6561)}{-2/3} = 4.5 \cdot \frac{6560}{6561} \cdot \frac{3}{2} = \frac{4.5 \cdot 3 \cdot 3280}{6561} = \frac{13.5 \cdot 3280}{6561} \approx 6.7499$. 5) $b_1 = -9, q = \sqrt{3}, n = 6$ $S_6 = \frac{-9((\sqrt{3})^6 - 1)}{\sqrt{3} - 1} = \frac{-9(27 - 1)}{\sqrt{3} - 1} = \frac{-9 \cdot 26}{\sqrt{3} - 1} = \frac{-234}{\sqrt{3} - 1}$. Избавимся от иррациональности: $S_6 = \frac{-234(\sqrt{3} + 1)}{3 - 1} = \frac{-234(\sqrt{3} + 1)}{2} = -117(\sqrt{3} + 1) = -117\sqrt{3} - 117$. 6) $b_1 = 8, q = -1/2, n = 4$ $S_4 = \frac{8((-1/2)^4 - 1)}{-1/2 - 1} = \frac{8(1/16 - 1)}{-3/2} = \frac{8(-15/16)}{-3/2} = -7.5 \cdot (-2/3) = 5$. ### Решение упражнения 871 1) $b_1 = 1, q = 2, n = 9$ $S_9 = \frac{1(2^9 - 1)}{2 - 1} = 512 - 1 = 511$. 2) $b_1 = 15, q = 2/3, n = 3$ $S_3 = \frac{15((2/3)^3 - 1)}{2/3 - 1} = \frac{15(8/27 - 1)}{-1/3} = \frac{15(-19/27)}{-1/3} = 15 \cdot \frac{19}{27} \cdot 3 = 15 \cdot \frac{19}{9} = \frac{5 \cdot 19}{3} = \frac{95}{3} = 31\frac{2}{3}$. 3) $b_1 = 18, q = -1/3, n = 5$ $S_5 = \frac{18((-1/3)^5 - 1)}{-1/3 - 1} = \frac{18(-1/243 - 1)}{-4/3} = \frac{18(-244/243)}{-4/3} = 18 \cdot \frac{244}{243} \cdot \frac{3}{4} = \frac{18 \cdot 3}{4} \cdot \frac{244}{243} = 13.5 \cdot \frac{244}{243} = \frac{27}{2} \cdot \frac{244}{243} = \frac{1 \cdot 122}{9} = \frac{122}{9} = 13\frac{5}{9}$. 4) $b_1 = 4, q = -\sqrt{2}, n = 4$ $S_4 = \frac{4((-\sqrt{2})^4 - 1)}{-\sqrt{2} - 1} = \frac{4(4 - 1)}{-\sqrt{2} - 1} = \frac{12}{-(1 + \sqrt{2})} = -\frac{12(\sqrt{2} - 1)}{2 - 1} = -12(\sqrt{2} - 1) = 12 - 12\sqrt{2}$.