Задание 44. Найдите сумму п членов геометрической прогрессии.

Давай вместе найдем сумму n членов геометрической прогрессии в каждом из предложенных случаев. Будем использовать формулу суммы n первых членов геометрической прогрессии: \[ S_n = \frac{b_1(q^n - 1)}{q - 1} \], где \( b_1 \) - первый член, \( q \) - знаменатель, \( n \) - количество членов. 1) \( b_1 = 2 \), \( b_2 = 6 \), \( q = 6:2 = 3 \), \( n = 4 \) \[ S_4 = \frac{2(3^4 - 1)}{3 - 1} = \frac{2(81 - 1)}{2} = 80 \] 2) \( b_1 = -2 \), \( b_2 = 4 \), значит, \( q = \frac{4}{-2} = -2 \), \( n = 8 \) \[ S_8 = \frac{-2((-2)^8 - 1)}{-2 - 1} = \frac{-2(256 - 1)}{-3} = \frac{-2(255)}{-3} = \frac{510}{3} = 170 \] 3) \( b_1 = 1 \), \( b_2 = -1 \), значит, \( q = \frac{-1}{1} = -1 \), \( n = 10 \) \[ S_{10} = \frac{1((-1)^{10} - 1)}{-1 - 1} = \frac{1(1 - 1)}{-2} = \frac{0}{-2} = 0 \] 4) \( b_1 = -1 \), \( b_2 = -5 \), значит, \( q = \frac{-5}{-1} = 5 \), \( n = 4 \) \[ S_4 = \frac{-1(5^4 - 1)}{5 - 1} = \frac{-1(625 - 1)}{4} = \frac{-624}{4} = -156 \] 5) \( b_1 = 25 \), \( q = \frac{1}{5} \), \( n = 3 \) \[ S_3 = \frac{25((\frac{1}{5})^3 - 1)}{\frac{1}{5} - 1} = \frac{25(\frac{1}{125} - 1)}{\frac{1 - 5}{5}} = \frac{25(\frac{1 - 125}{125})}{\frac{-4}{5}} = \frac{25(\frac{-124}{125})}{\frac{-4}{5}} = \frac{\frac{-124}{5}}{\frac{-4}{5}} = \frac{-124}{5} \cdot \frac{5}{-4} = 31 \] 6) \( b_1 = 128 \), \( q = 0.5 = \frac{1}{2} \), \( n = 7 \) \[ S_7 = \frac{128((\frac{1}{2})^7 - 1)}{\frac{1}{2} - 1} = \frac{128(\frac{1}{128} - 1)}{\frac{1 - 2}{2}} = \frac{128(\frac{1 - 128}{128})}{\frac{-1}{2}} = \frac{128(\frac{-127}{128})}{\frac{-1}{2}} = \frac{-127}{\frac{-1}{2}} = -127 \cdot (-2) = 254 \] 7) \( b_1 = \frac{49}{4} \), \( q = -\frac{2}{7} \), \( n = 3 \) \[ S_3 = \frac{\frac{49}{4}((-\frac{2}{7})^3 - 1)}{-\frac{2}{7} - 1} = \frac{\frac{49}{4}(-\frac{8}{343} - 1)}{-\frac{2}{7} - \frac{7}{7}} = \frac{\frac{49}{4}(-\frac{8 + 343}{343})}{-\frac{9}{7}} = \frac{\frac{49}{4}(-\frac{351}{343})}{-\frac{9}{7}} = \frac{\frac{49}{4}(-\frac{351}{343})}{-\frac{9}{7}} = \frac{\frac{7}{4}(-\frac{351}{49})}{-\frac{9}{7}} = \frac{-\frac{351}{28}}{-\frac{9}{7}} = -\frac{351}{28} \cdot -\frac{7}{9} = \frac{351}{4} \cdot \frac{1}{9} = \frac{39}{4} = 9.75 \] 8) \( b_2 = -0.25 = -\frac{1}{4} \), \( q = 0.5 = \frac{1}{2} \), тогда \( b_1 = \frac{b_2}{q} = \frac{-\frac{1}{4}}{\frac{1}{2}} = -\frac{1}{4} \cdot 2 = -\frac{1}{2} = -0.5 \), \( n = 4 \) \[ S_4 = \frac{-0.5((\frac{1}{2})^4 - 1)}{\frac{1}{2} - 1} = \frac{-0.5(\frac{1}{16} - 1)}{-\frac{1}{2}} = \frac{-0.5(\frac{1 - 16}{16})}{-\frac{1}{2}} = \frac{-0.5(\frac{-15}{16})}{-\frac{1}{2}} = \frac{\frac{15}{32}}{-\frac{1}{2}} = \frac{15}{32} \cdot -2 = -\frac{15}{16} = -0.9375 \] 9) \( b_2 = -0.2 \), \( q = -2 \), тогда \( b_1 = \frac{b_2}{q} = \frac{-0.2}{-2} = 0.1 \), \( n = 7 \) \[ S_7 = \frac{0.1((-2)^7 - 1)}{-2 - 1} = \frac{0.1(-128 - 1)}{-3} = \frac{0.1(-129)}{-3} = \frac{-12.9}{-3} = 4.3 \]

Ответ:

• 1) 80
• 2) 170
• 3) 0
• 4) -156
• 5) 31
• 6) 254
• 7) 9.75
• 8) -0.9375
• 9) 4.3

Ты молодец! У тебя всё получится!