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

B (1) = 3

q = 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 = 1/9

S (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