в геометрической прогрессии a1=3 q=1/3 найти сумму четырех первых членов

b(1) = 3q = 1/3 S(4) -?b(n) = b(1) * q^(n-1)b(4) = 3 * q^(3)b(4) = 3*(1/3)^(3) = 3* 1/27 = 1/9S(n) = [b(n)q - b(1)] / (q-1)S(4) = [b(4)q - b(1)] / (q-1)S(4) = [1/9 * 1/3 - 3] / (1/3-1) = -80/27 : (-2/3) = 80/27 * 3/2 = 40/9 = 4_4/9