# Sum of the terms of a geometric sequenceVerify if the sum 4^n - 1 represents the sum of the terms of a geometric sequence.

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### 2 Answers

Let 4^n - 1 represent the sum of m terms of a GP. 4^(n - 1) - 1 is the sum of m - 1 terms of the GP.

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

=> 4^n - 4^n/4

For two consecutive terms Tn and Tn+1

Tn+1 / Tn = [4^(n + 1) - 4^(n + 1)/4]/(4^n - 4^n/4)

=> [4*4^n - 4*4^n/4]/(4^n - 4^n/4)

=> 4*(4^n - 4^n/4)/(4^n - 4^n/4)

=> 4

**As we arrive at a common ratio between consecutive terms, 4^n - 1 represents the sum of terms of a GP.**

Finding the general term bn of the sequence, allows us to determine any 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)

Now, we'll calculate 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

According to 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 represents the sum of the terms of a geometrical progression!**