Respuesta :

a_{1} = 4
a_{2} = 12

a_{2} = a_{1}*q
12 = 4q
q = 3

S20 = a_{1}*[tex] \frac{(q^{20}-1)}{(q-1)} [/tex]

S20 = 4*[tex] \frac{(3^{20}-1)}{3-1} [/tex]

S20 = 4*[tex] \frac{3486784401-1}{2} [/tex]

S20 = 2*3486784400

S20 = 6,973,568,800