Question

calcular a soma dos onze primeiros termos da P.G (1\32^2 1/16^2 1/8^2)?

Answer 1

a1 =  ( 1/32)² = 1/1024
a2 = (1/16)² = 1/256
a3 =( 1/8)²   = 1/64

q = ( 1/16)² : ( 1/32)²  =  ( 1/16)²  * ( 32/1)² = ( 32/16)²   = 2²  = 4 ***
ou
q = 1/256 :  1/1024 = 1/256 * 1024/1 = 1024/256 = 4 ***
n = 11 
S11 =  1/1024 ( 4¹¹  - 1)/ ( 4 - 1)
S11 = 1/1024 ( 4194304 - 1)/ 3
S11 = 1/1024 ( 4194304)/3
S11 = 1/1024 . 4194304/3 = 4194304/3072   = 4096/3