Question

Fie (Bn) o progresie geometrică. Să se determine suma Sn a primilor n termeni dacă:
a) b1= 3, q=2,n=6
b) b1= 2,5, q =1,5 ,n=5
c) b1=8, b3=2, n=8
d) b1 = -1,n = 100,q = -1.
Vă rog, ajutaţi-mă!
Dau coroana !

Answer 1

Sn = b1 * (q^n-1)/q-1
a) S6 = 3*(2^6-1)/(2-1) = 3*64/1 = 192

b) S5 = 2,5 * [(1,5)^5-1]/(1,5-1) = (2+5/10) *[(3/2)^5-1]/(3/2-1) = (5/2) * (243/32-1)/(1/2) = (5/2) * (211/32)/(1/2) = (5/2) * (211/32)*(2/1) = 1055/32

c) bn = b1*q^(n-1)
    b3 = b1*q^2 = 8*q^2 = 2
    q^2 = 2/8 = 1/4
    q = 1/2
   S8 = 8*[(1/2)^8-1]/(1/2-1) = 8*(1/256-1)/(-1/2) = 8*(-255/256)/(-1/2) = 8*(255/256)*(2/1) = 255/16

d) S100 = (-1)*[(-1)^100 - 1]/(-1-1) = (-1)*(1-1)/(-2) = (-1)*0/(-2) = 0