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Math Help - Infinite Sum of a Geometric Series

  1. #1
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    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.
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  2. #2
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    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}.
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  3. #3
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    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:

    <br />
\frac {3(1-0.15^9)} {1-0.15}
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  4. #4
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    Quote Originally Posted by Spudwad View Post
    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:

    <br />
\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}
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  5. #5
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    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:

    <br />
\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.
    Last edited by Spudwad; February 23rd 2010 at 04:44 PM.
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