# Geometric Sequence.How to check if a given sum represents the sum of the terms of a geometric sequence?

### 2 Answers | Add Yours

The sum of n terms of a geometric sequence is given as a*(r^n - 1)/(r - 1), where r is the common ratio and a is the first term

4^n - 1

=> (2^(2n) - 1) / (2 - 1)

Here a = 1 and r = 2. But 2n can only take on even values.

The expression 4^n - 1 can be the sum of terms of a geometric sequence but it holds only when the number of terms is even.

We'll establish a numerical expression for the sum of n terms:

S, = (4^n)-1

We'll determine the general term bn, and then, 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

We'll eliminate like terms and we'll factorize by 4^n:

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

Since we know the general term bn, we'll compute the first 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 of a geometric sequence, we'll verify if

b2=sqrt (b1*b3)

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

3*4=3*4

**Sn = 4^n - 1 is the sum of the n terms of a geometric sequence.**