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Math Help - Geometric Series....got it down to algebra...but i can figure it out

  1. #1
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    Geometric Series....got it down to algebra...but i can figure it out

    so it was

    If 1+x+x^2+x^3+.....+x^24=10 what is x

    so i got

    ((x^25) - x) / (x -1) = 9

    and i put in my calc and got the right answer..but how can i do it just by solving the algebra?

    the answer is .90927693 (almost positive)
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  2. #2
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    another one

    also

    1+x+x^2+x^3........=10

    set it up: (0-x)/(x-1) = 9 so x = .9

    but i did it using my calc also

    im clearly missing a big piece of algebra knowledge
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by stones44 View Post
    so it was

    If 1+x+x^2+x^3+.....+x^24=10 what is x

    so i got

    ((x^25) - x) / (x -1) = 9

    and i put in my calc and got the right answer..but how can i do it just by solving the algebra?

    the answer is .90927693 (almost positive)
    \sum_{n = 0}^kar^n = a \cdot \frac{1 - r^{k + 1}}{1 - r}

    \frac{1- x^{25}}{1 - x} = 10
    (Note the difference between my formula and yours. I took a0 = 1 and n = 24. You took a0 = x and n = 24. Clearly your n should have been 23.)

    So:
    1 - x^{25} = 10(1 - x)

    1 - x^{25} = 10 - 10x

    x^{25} - 10x + 9 = 0

    You want to find an exact root of this equation? Can't be done in general. There are no rational roots, at least, so we are stuck with numerical approximation.

    -Dan
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by stones44 View Post
    also

    1+x+x^2+x^3........=10

    set it up: (0-x)/(x-1) = 9 so x = .9

    but i did it using my calc also

    im clearly missing a big piece of algebra knowledge
    This one, ironically, we can solve.

    \sum_{n = 0}^{\infty}ar^n = \frac{a}{1 - x}

    So
    1 + x + x^2 + ~...~ = \frac{1}{1 - x} = 10

    1 = 10(1 - x)

    1 = 10 - 10x

    10x = 9

    x = \frac{9}{10}

    -Dan
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    arent you supposed to do last *multipler - first / multipler - 1

    and the multipler is X and first is x because 1 is like a seperate thing (not part of the series)
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by stones44 View Post
    arent you supposed to do last *multipler - first / multipler - 1

    and the multipler is X and first is x because 1 is like a seperate thing (not part of the series)
    \sum_{n = 0}^{k}ar^n = a \cdot \frac{1 - r^{k + 1}}{1 - r}

    Here you have
    \sum_{n = 0}^{24} 1 \cdot x^k = 1 + x + x^2 + ~ ... ~ + x^{23} + x^{24}

    So a = 1, the coefficient of x^k in the sum, otherwise known as the first term in the summation, and r = x, the geometric ratio.

    What you tried to do was this:
    1 + (x + x^2 + ~...~ + x^{23} + x^{24} ) = 10

    x + x^2 + ~...~ + x^{23} + x^{24} = 9

    There is nothing wrong in doing this, but now your series is
    \sum_{n = 0}^{23}x \cdot x^k = x \cdot \frac{1 - x^{23}}{1 - x}
    so a = x and r = x.

    -Dan
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