What is the value of r of the geometric series? \(\sum_{n=1}^{3}\)(1.3)(0.8)^n-1.

Solution:

The geometric series is summarized as 

\(\sum_{n=1}^{3}\)(1.3)(0.8)^n-1.

The series generated by the above expression will be as follows:

(1.3)(0.8)^1-1 + (1.3)(0.8)^2-1 + (1.3)(0.8)^3-1

= (1.3)(0.8)⁰   +  (1.3)(0.8)¹ +  (1.3)(0.8)²

 = (1.3) + (1.3)(0.8) + (1.3)(0.8)²

From the three terms given above we can determine the common ratio r as

r = (1.3)(0.8)¹/(1.3) = (1.3)(0.8)²/(1.3)(0.8)¹  = (0.8)

Hence the common ratio of the geometric series is 0.8

What is the value of r of the geometric series? \(\sum_{n=1}^{3}\)(1.3)(0.8)^n-1.

Summary:

 The value of r of the geometric series \(\sum_{n=1}^{3}\)(1.3)(0.8)^{n - 1} is 0.8.