Question

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

Answer 1

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