# Prove that the series that have the sum of the n terms Sum=4^n -1, is a geometric progression.

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### 2 Answers

If we want to prove that the given series is a geometrical progression, we'll have to substantiate that having 3 consecutive terms of the progression, the middle one is the geometric mean of it's neighbors.

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

From enunciation:

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

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

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

We'll express 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 enunciated above, we'll verify if

b2=sqrt (b1*b3)

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

**3*4=3*4**

Sn is a geometrical progression!

To prove that sum = 4^n-1 is asum of the first n terms of ageometric series.

Let Sn = 4n^2-1.

The first term of the series a1 = s1 = 4^1-1 = 3, where a1 is the first term

S2 = 4^4-1= 16-1 = 15.

But S2 = a1+a2 = a1+ a1*r, where r is the ratio of the second term to first term.

a1+a1r = S2 = 15.

3+3r = 15.

So r = (15-3)/3 = 4.

Therefore let create a geometric series with a1 = 3as the first term and r = 4as common ratio and find the sum to n terms: S(n) = 3+3*4+3*4^2+3*4^3+...3*4^(n-1)

S(n) = 3{1+ 4+4^2+4^3+.....4^(n-1)}

S(n) = 3{4^n-1}/(4-1).

S(n) = 4^n - 1.

Therefore S(n) =4^n-1 = S(n), is the sum of n terms of a GP.