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Math Help - Infinite Series

  1. #1
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    Infinite Series

    I originally thought this Infinite Series would be fairly easy, however I can't figure out how to evaluate it.


    infinity
    ∑ (2^(j+4)) / (7^(j-3))
    j=6


    is there a certain formula i need to follow?
    Thanks.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by stevo25 View Post
    I originally thought this Infinite Series would be fairly easy, however I can't figure out how to evaluate it.


    infinity
    ∑ (2^(j+4)) / (7^(j-3))
    j=6


    is there a certain formula i need to follow?
    Thanks.
    \sum_{j=6}^{\infty}p^{j+4}{q^{j-3}}=p^4 q^3\sum_{j=6}^{\infty}\left(\frac{p}{q}\right)^j. Look familiar?
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  3. #3
    MHF Contributor chiph588@'s Avatar
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    Champaign, Illinois
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    Quote Originally Posted by stevo25 View Post
    I originally thought this Infinite Series would be fairly easy, however I can't figure out how to evaluate it.


    infinity
    ∑ (2^(j+4)) / (7^(j-3))
    j=6


    is there a certain formula i need to follow?
    Thanks.
     \sum_{j=6}^\infty \frac{2^{j+4}}{7^{j-3}} = 2^4\cdot7^3\sum_{j=6}^\infty \frac{2^{j}}{7^{j}} = 2^4\cdot7^3\sum_{j=6}^\infty\left(\frac27\right)^j

    Now we're left with a geometric series. Can you take it from here?

    EDIT: Ah! Drexel28 beat me to the chase!
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  4. #4
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    Ah, it is pretty simple! After that you just add 6 to the j to set the index to 0 and then put (2/7)^6 outside the summation. Then you can plug it into

    (ar^(n+1) - a) / (r-1)

    formula and take the limit as it approaches infinity.

    Thanks!
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