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Math Help - Write the series using summation notation

  1. #1
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    Write the series using summation notation

    3. a) Write the series 60 -15 +15/4 -15/16 +15/64 -15/256.

    My work:

    I discovered each term is divided by -4.

    60 / -4 = -15
    15 / -4 = 15/4
    15/4 / -4 = -15/16

    Total terms 6 (above sigma, also n)
    First term k=1 (first term)

    Now for the part on the right of sigma, I have difficulty with this.

    I wrote the part on the right of sigma, which should be {a+(n-1)d}, as {60+95) / -4}.

    I've never encountered a series where everything is divided rather than multiplied so I'm a little lost.

    Please help.
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by thekrown View Post
    3. a) Write the series 60 -15 +15/4 -15/16 +15/64 -15/256.

    My work:

    I discovered each term is divided by -4.

    60 / -4 = -15
    15 / -4 = 15/4
    15/4 / -4 = -15/16

    Total terms 6 (above sigma, also n)
    First term k=1 (first term)

    Now for the part on the right of sigma, I have difficulty with this.

    I wrote the part on the right of sigma, which should be {a+(n-1)d}, as {60+95) / -4}.

    I've never encountered a series where everything is divided rather than multiplied so I'm a little lost.

    Please help.
    This is geometric...

    \sum_{n=0}^{\infty}60\left(\frac{-1}{4}\right)^n
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  3. #3
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    My question only lists 6 terms for the series. Would it be okay to list n=1 and in the right side of sigma have n-1 instead of just n?

    Second, with a 6 term series would the infinity sign then be 6?
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by thekrown View Post
    My question only lists 6 terms for the series. Would it be okay to list n=1 and in the right side of sigma have n-1 instead of just n?

    Second, with a 6 term series would the infinity sign then be 6?
    Sorry...

    Correction

    \sum_{n=1}^660\left(\frac{-1}{4}\right)^{n-1}
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  5. #5
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    Thank you. Really helped a lot. Cheers!
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  6. #6
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    I just reviewed this question and get -4 as the 'r' value not -1/4. Can you help clarify this issue?
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