Geometric Series: What is the ratio, r?

[MatheMatix]

Hello,

I'm stuck on a homework problem. I already know the answer. But I'm trying to figure out how to solve it.

From Thomas' Calculus Early Transcendentals Section 11.2
Series with Geometric Terms.
"In Exercises 7-14, write out the first few terms of each series to show how the series starts. Then find the sum of the series."
Exercise 11

ASSUMPTIONS:
I am assuming this is a geometric series, i.e. a + ar + ar^2 + ... + ar^ (n-1) + ...
This is the first problem I've encountered where there are two "terms" or fractions in the summation, and I think I'm having problems with that in particular. (I had little trouble solving the previous problems with only one term in their summation in this section.)

And to find where a geometric series converges, we use the formula a / (1 - r) provided | r | < 1. ( If| r | >= 1, the series diverges. )

ANSWER:
The first few terms of the series is
(5 + 1) + (5/2 + 1/3) + (5/4 + 1/9) + (5/8 + 1/27) + ...

And the sum is 23/2.

QUESTION:
My problem is finding out what r is.

As I understand it, a is supposed to be 6 (the first term of the series) and r, a fixed ratio, is supposed to be a fraction raised to the nth power times a that will give me the next terms and allow me to find the sum/where it converges by using the above formula. However, looking at the first four terms in the summation, I do not detect any pattern to help me find what r is, given a is 6...

Thank you in advance.

 

[stapel]

Are you sure you're supposed to find one geometric series for this...? Instead, I would go this way:


. . . . .\(\displaystyle \L \sum_{n=0}^{\infty}\, \left(\frac{5}{2^n}\, +\, \frac{1}{3^n}\right)\)


. . . . .\(\displaystyle \L \sum_{n=0}^{\infty}\, \frac{5}{2^n}\, +\, \sum_{n=0}^{\infty}\, \frac{1}{3^n}\)


. . . . .\(\displaystyle \L 5\, \sum_{n=0}^{\infty}\, \frac{1}{2^n}\, +\, \sum_{n=0}^{\infty}\, \frac{1}{3^n}\)


Then find the two sums (with r = 1/2 and r = 1/3, respectively), multiply the first sum by 5, and then add the two values.

Eliz.

 

[MatheMatix]

Thanks very much for your post. That's correct. I realized that was the way to do it soon after posting!