# SumThe sum of n terms is 4^n - 1. Prove it is a geometric series!

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You need to use the equation that gives the summation of terms of geometric series, such that:

`Sigma_(k=1)^n b_k = b_1*(q^n - 1)/(q - 1)`

The problem provides the information that `Sigma_(k=1)^n b_k = 4^n - 1` , such that:

`b_1*(q^n - 1)/(q - 1) = 4^n - 1`

Considering `q^n - 1 = 4^n - 1` yields:

`q^n - 1 = 4^n - 1 => q^n = 4^n => q = 4`

Since `b_1*(4^n - 1)/(4 - 1) = 4^n - 1 => b_1 = 4 - 1 => b_1 = 3` .

**Hence, checking what is the geometric series whose summation of its terms yields `4^n - 1,` yields that the first term of series is `b_1 = 3` and its common ratio is **`q = 4.`

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.