How to Find the Sum of a Geometric Series with Variables?

[Spencer23]

Hey,

Sorry if I am in the wrong part of the forums not sure where this question goes. I am having trouble with a geometric series that has letters involved. I understand the forumla for finding the sum of first n elements with just numbers. However the series i have is ..

a1 = -5, a2 = -5x, a3 = -5x^2...

How do i go about finding the sum of the first 8 elements with the normal formula for doing so? Which I am under the impression is ...

a1(1-r^n)/1-r

 

[I like Serena]

Hey,

Sorry if I am in the wrong part of the forums not sure where this question goes. I am having trouble with a geometric series that has letters involved. I understand the forumla for finding the sum of first n elements with just numbers. However the series i have is ..

a1 = -5, a2 = -5x, a3 = -5x^2...

How do i go about finding the sum of the first 8 elements with the normal formula for doing so? Which I am under the impression is ...

a1(1-r^n)/1-r

Hi Spencer23! Welcome to MHB! :)

You are entirely correct.
So with $a_1=-5$, $n=8$, and $r=x$, we get:
$$a_1 + a_2 +...+a_8 = -5 \cdot \frac{1-x^8}{1-x}$$

 

[MarkFL]

If you wanted to work the problem without a formula, you could state:

\(\displaystyle S=-5-5x-5x^2-5x^3-5x^4-5x^5-5x^6-5x^7=-5\left(1+x+x^2+x^3+x^4+x^5+x^6+x^7\right)\)

Now, multiply both sides by $x$:

\(\displaystyle Sx=-5\left(x+x^2+x^3+x^4+x^5+x^6+x^7+x^8\right)\)

If we subtract the first equation from the second, we obtain:

\(\displaystyle S(x-1)=-5\left(x^8-1\right)\)

Hence:

\(\displaystyle S=-5\frac{x^8-1}{x-1}=-5\frac{1-x^8}{1-x}\)

 

[burgess]

Nice answer!

If you wanted to work the problem without a formula, you could state:

\(\displaystyle S=-5-5x-5x^2-5x^3-5x^4-5x^5-5x^6-5x^7=-5\left(1+x+x^2+x^3+x^4+x^5+x^6+x^7\right)\)

Now, multiply both sides by $x$:

\(\displaystyle Sx=-5\left(x+x^2+x^3+x^4+x^5+x^6+x^7+x^8\right)\)

If we subtract the first equation from the second, we obtain:

\(\displaystyle S(x-1)=-5\left(x^8-1\right)\)

Hence:

\(\displaystyle S=-5\frac{x^8-1}{x-1}=-5\frac{1-x^8}{1-x}\)

 

[CellCoree]

Hi there,

I can provide some guidance on how to approach this problem. First, it's important to understand the concept of a geometric progression. In a geometric progression, each term is found by multiplying the previous term by a constant value, called the common ratio. In your case, the common ratio is -5x.

To find the sum of the first n terms in a geometric progression, there are a few different formulas you can use. One option is the formula you mentioned, which is a1(1-r^n)/1-r. In this formula, a1 represents the first term in the series, r represents the common ratio, and n represents the number of terms you want to sum.

In your case, a1 = -5, r = -5x, and n = 8. So, you can plug these values into the formula and solve for the sum of the first 8 terms.

Another option is to use the formula Sn = a1(1-r^n)/(1-r), where Sn represents the sum of the first n terms. In this formula, a1 and r have the same meaning as before, but the value of n is one less than the number of terms you want to sum. In your case, n = 7.

I hope this helps and let me know if you have any further questions. Good luck with your problem!