Question

найдите сумму первых семи членов геометрической прогрессии (bn), если b3=64, q=1/2

Answer 1

b3 = b1*q^2 ==>

b1*q^2 = 64
b1*(1/2)^2 = 64
b1 = 64 / 0.25
b1 = 256

S7 = (b1*(1 - q^7))/(1 - q) = (256*(1 - (1/2)^7))/(1 - (1/2)) = 508