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

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

Given:

Sn = 4^n -1

We need to determine wether n are terms of G.S

We know that the sum of a geometric series is:

Sn = a1+a2 + ....+ an

==> a1+a2+...+ an = 4^n - 1

==: sn = 4^n -1 - a1-a2-a3 - ...- a(n-1)

= 4^n -1 - (a1+a2+...+a(n-1)

but we know that:

(a1+a2+...+a(n-1) = S(n-1)= 4^(n-1)  -1

==> Sn = 4^n -1 - (4^(n-1) - 1

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

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

= 4^n(1- 1/4)

= 4^n(3/4)

= 3*4^(n-1)

==> Sn = 3*4^(n-1)   which s a sum of a G.S with first terms is 3 and common difference (r)= 4

==> a1= 3

=> a2= 3*4= 12

a3= 3*4^2 = 48

.....

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

The sum of the series 4^n -1.

Therefore S1 = 4^1 -1 = 3. Therefore a= a1 = 3,

S2 = 4^2 -1  = 15. Therefore a1+a2 = 15. So  a2 = 15- a1 = 15-3 = 12.

a3 = S3-S2 = 4^3-1 - (4^2 -1) = 48

a4 = S4-S3 = (4^4-1) -(4^31) = 192.

Thus the Geometric series is , (a1, a2,a3 ,a4,...) = (3 , 12, 48 , 192,...)

Or the GP is a = 3, common ratio  r = 4 , and the sum Sn = a(r^n-1)/(4-1) = 3(4^n-1)(4-1) = 4^n-1.

Thus the series whose  sum of the n terms of the series is 4^n-1 is a GP with starting term 3 and common ratio 4.

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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.

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