S = 4^n -1

Let us determine the terms:

s1= 4^0 = 1

S2= 4^1 = 4

S3= 4^2 = 16

S4 = 4^3 = 64

......

We notice that:

1, 4, 16, 64 , ..... is a Geometric progression with a constant difference r = 4 such that:

s1= 1

s2= 1*4

s3= 1*4^2

s4= 1*4^3

.......

sn = 1*4^n-1 = 4^n-1

Then S is a G.P

The give series has the sum s = 4^n-1.

To prove that this is a GP.

Solution:

Sn = 4^n -1 by data

Replace n by n-1 and we get:

Sn-1 = 4^(n-1)-1 by data

Sn = Sn-1 +an. Or

an = Sn-Sn-1 = (4^n -1) - [4^(n-1) - 1] = 4^n - 4^(n-1) = 4^(n-1)* (4-1) = 3*4^(n-1)

an = 3*4^(n-1).

But an = a* r^(n-1) is the nth term of a GP with starting term a1 =a and common ratio r.

Therefore identifying a*r^(n-1) with 3*4^(n-1) , we get a1 = a = 3 and r = 4 .

So the GP with the beginning term 3 and common ratio 4 has the sum to n terms Sn = 4^n - 1.

Let's recall that for 3 consecutive terms of a geometric progression, the middle one 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

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

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

But (b1+b2+b3+...+b(n-1))=S(n-1)=[4^(n-1)]-1, is the sum of the first (n-1) terms.

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

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

We'll write the 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**

**The series of n terms is a geometric progression.**