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 formula 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 formula 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:

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

Now, multiply both sides by $x$:

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

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

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

Hence:

$$S=-5\frac{x^8-1}{x-1}=-5\frac{1-x^8}{1-x}$$

 

[burgess]

Nice answer!

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

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

Now, multiply both sides by $x$:

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

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

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

Hence:

$$S=-5\frac{x^8-1}{x-1}=-5\frac{1-x^8}{1-x}$$