The sum of n terms is 4^n - 1. Prove that is a geometric series.

Let's recall the rule of 3 consecutive terms of a
geometric progression, where the middle term is the geometric mean of the ones adjacent
to it.

We'll determine the formula of the general term bn,
and after finding it, we'll utter any other term of the
progression.

From enunciation, the sum of n
terms:

Sn = b1+b2+b3+...+bn
(1)

But the result of the sum is 4^n - 1. We'll substitute
in (1)

(4^n)-1=
b1+b2+b3+...+bn

We'll subtract b1+b2+b3+...+b(n-1) both
sides:

bn =
(4^n)-1-(b1+b2+b3+...+b(n-1))

But (b1+b2+b3+...+b(n-1)) is
the sum of the first (n-1) terms.

We'll put the sum of n-1
terms as
S(n-1)=[4^(n-1)]-1.

bn=(4^n)-1-4^(n-1)+1

bn=4^n(1-1/4)=4^n*3/4=3*4^(n-1)

We'll compute 3
consecutive terms,
b1,b2,b3.

b1=3*4^0

b2=3*4^(2-1)=3*4

b3=3*4^(3-1)=3*4^2

Following
the rule:

 b2=sqrt
(b1*b3)

3*4=
sqrt(3*1*3*16)

3*4=3*4

Since
the computed terms were chosen randomly, the series of n terms is a geometric
progression.