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Math Help - Find the sums:

  1. #1
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    Find the sums:

    5 + 15 + 45 + .... + 98, 415
    I think i know how to do this but i dont know the pattern to the numbers and the comma at the end threw me off.
    Any help appreciated. THx.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ijleyton View Post
    5 + 15 + 45 + .... + 98, 415
    I think i know how to do this but i dont know the pattern to the numbers and the comma at the end threw me off.
    Any help appreciated. THx.
    I think the last number is 98415. Some books still use a comma to separate powers of 3 in the number. (ie. 1000 --> 1,000 and a million is 1,000,000.)

    It looks like a geometric series:
    a_k = 5*3^{k - 1}

    The sum of a geometric series (with the first term equal to 1) is
    S_n = (1 - r^{n+1})/(1 - r) = r^0 + r^1 + ... + r^n
    where, in this case, r = 3 and n is the label of the last term in the series.

    So what is n?

    98415 = 5*3^{n - 1}

    19683 = 3^{n - 1}

    n - 1 = log_3(19683) = ln(19683)/ln(3) = 9

    n = 10

    So
    S_10 = (1 - 3^{10 + 1})/(1 - 3) = 88573

    This is the sum for
    S_10 = 3^0 + 3^1 + 3^2 + ... + 3^{10}

    We need to remember to multiply by 5, so our sum is:
    5*88573 = 442865

    -Dan
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