Is the queue (Sn), a geometrical progression, where Sn is the sum of n terms of progression; Sn=(4^n) -1

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giorgiana1976's profile pic

Posted on

One of the methods to demonstrate that we deal with a geometrical progression is to substantiate that having 3 consecutive terms of the progression, the middle one is the geometric average of the ones adjacent to it.

First of all, we'll try to find 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!

neela's profile pic

Posted on

Sn = a+ar+ar^2+....ar^n-1= a(r^n  -1)/(r-1) = 4^n-1

comparing,we get

ar^n/(r-1) =4^n  -a/(r-1) = -1=> a=r-1

(ar^n  -1)/(r-1)= 4^n=> r=4  a =3

Therefore, the required G P is:

Sn = 3+3*4+3+4^2+3*4^3+.....+3*4^(n-1).

 

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