дана геометрическая прогрессия 1 1 2 1 4 1 8 найдите s8 sn

b1 = 1
b2 = 1/2
q = b2/b1 = 1/2 : 1 = 1/2
S₈ = b1 * ([tex]q^{8} [/tex]-1)/(q-1) = 1*([tex] ( \frac{1}{2}) ^{8} [/tex]-1)(1/2-1) = 
 = (1/256 - 1)/(1/2-1) = 255/128 = 1 127/128
Sn = b1 * (qⁿ-1)/(q-1) = 1*((1/2)ⁿ-1)/(1/2-1) = (1/2)ⁿ-1/(-1/2) = (1-2ⁿ)/2ⁿ : (-1/2) = 
(1-2ⁿ)/2ⁿ * (-2) = - (1-2ⁿ)/ (2ⁿ⁻¹)