# Math Help - Infinite Sum of a Geometric Series

1. ## Infinite Sum of a Geometric Series

I am having a bit of difficulty in determining the sum of an infinite geometric series:

Find the sum (assume ):
$2 + 4x + 8x^2 + 16x^3 +32x^4 + ...$

I was shown that the sum of this series is equal to the following when :

$\frac {2} {1-x}$

Where 2 is the first value. But I thought that since this series was increasing by a common ratio of 2x that it really depends upon how small or large x really is. This is where I am stuck. I am just not sure how to go around the lack of a definitive x value. I appreciate the help.

2. The series $2 + 4x + 8x^2 + 16x^3 + 32x^4 + ...$ can be written

$2 + 2(2x) + 2(2x)^2 + 2(2x)^3 + 2(2x)^4 + ...$

which is clearly a geometric series with first term 2 and a common ratio of 2x.

Therefore, the sum is

$\frac{2}{1 - 2x}$.

3. Ah, woops. Thank you, I just didn't realize that you could leave this in terms of X. I appreciate that, but I also have one more question, sort of revolving around the same concept, so I will post it below.

Find the sum of:

$3(.015)^4 + 3(0.15)^5 + 3(0.15)^6 + ... +3(0.15)^{12}$

Would the sum of this simply be:

$
\frac {3(1-0.15^9)} {1-0.15}$

Ah, woops. Thank you, I just didn't realize that you could leave this in terms of X. I appreciate that, but I also have one more question, sort of revolving around the same concept, so I will post it below.

Find the sum of:

$3(.015)^4 + 3(0.15)^5 + 3(0.15)^6 + ... +3(0.15)^{12}$

Would the sum of this simply be:

$
\frac {3(1-0.15^9)} {1-0.15}$
Almost. However, the first term is not 3; it is $3(0.15)^4$.
Therefore, the sum is

$\frac{3(0.15)^4(1 - 0.15^9)}{1 - 0.15}$

5. I just wanted to say thanks to everyone that has helped me out. I had missed class yesterday and this has helped make it very clear. Thanks again.

EDIT:
Sorry for asking so many questions about this, I just have two more questions regarding closed form formulas for geometric series and summing infinite series.

For the closed form formula, I have an expression for the $nth$ term:

$7(7/8)^n$

But I am not sure how to convert this to closed form since I am not really sure what closed form looks like. Is it just to convert it into a finite sum equation like this:

$
\frac{7(1-\frac {7} {8}^n)} {(1- \frac {7} {8})}$

For the infinite series, I have the formula:

$\frac {1} {4} \sqrt{h}$

Where $h = 7(7/8)^n$

So the very first term would be:

$\frac {1} {4} \sqrt{7\frac {7} {8}}$

And this would go in the numerator, but I am not sure what would go in the denominator. Again, I really appreciate the help thus far.