Помогите решить геометрическую прогрессию. b1=3, q = 1/2, S6-?

Sn = (bn * q - b1) / (q - 1)

bn = b1 * q^ (n-1) (q в степени n-1) откуда b1 = bn / q^ (n-1)

|bn| = корень (bn-1 * bn+1)

q = bn+1 / bn

1) b7 = b1 * q^ (7-1) = 2 * 3^6 = 2 * 729 = 1458

S7 = (b7 * q - b1) / (q - 1) = (1458 * 3 - 2) / (3 - 1) = 4372 / 2 = 2186

2) b1 = b6 / q^ (6-1) = 2,56 / 2^5 = 0,08

3) b2 = корень (b1 * b3) = корень (4 * 36) = корень (144) = 12

q = b2 / b1 = 12 / 4 = 3

b4 = b3 * q = 36 * 3 = 108

S4 = b1 * (1 - (-5) ^4) / (1 - (-5))

-416 = b1 * (1 - 625) / 6

b1 = - 416 / - 104 = 4

b6 = b1 * q^ (6-1) = 4 * (-5) ^5 = - 12500

S6 = (b6 * q - b1) / (q - 1) = (-12500 * (-5) - 4) / (-5 - 1) = 62496 / - 6 = - 10416

b5 = b4 * q = 108 * 3 = 324 Sn = b1 * (1 - q^n) / (1 - q) при q не равном 1