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

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

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The sum of n terms of the series is given by 4^(n -1)

The sum of n + 1 terms is 4^(n + 1 - 1) = 4^n

The sum of n - 1 terms is 4^(n - 1 - 1) = 4^n/16

We see that the (n - 1)th term is 4^n/16, the nth term is 4^n/4 and the (n + 1)th term is 4^n.

4^n/(4^n/4) = (4^n/4)/(4^n/16) = 4

Each of the terms has a common ratio of 4.

This proves that the series is a geometrical progression

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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!

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