Entonces 2d q 4 = 20(1)4d q 2 = 12(2).
(1)* 2-(2)(2q^2 7)(q^2-4)=0
∵q gt; ∴q=2 reemplaza d=2.
an = 1 2(n-1)= 2n-1
bn=2^(n-1)
(2)Sn=1 3 /2 5/2^2 .... (2n-1)/2^(n-1) (3)
2Sn=2 3 5/2 ..... (2n-1 )/2^(n-2) (4)
(4)-(3) Sn=2 2 2/2 ... 2/2^(n-2)-(2n-1 )/2^(n-1)
=4 [1 1/2 ... 1/2^(n-1)]-(2n-1)/2^(n-1)
=4 [1-1/2^(n-1)]/(1-1/2)-(2n-1)/2^(n-1)
=4 2-2/2^(n-1)-(2n-1)/2^(n-1)
=6-(2n 1)/2^(n-1)