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Math Help - [SOLVED] Geometric Series Word Problem

  1. #1
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    [SOLVED] Geometric Series Word Problem

    I now how to workout this problem, but I make a few small errors along the way and I need some sort of checking or help.

    Problem.

    A ball is drooped from a height of 10 feet (Assume no air resistance), each bounce is (\frac{3}{4}) of the height of the bounce before. Thus, after the balls hit the floor for the first time, the ball rises to a height of 10(\frac{3}{4}), after the second time, 10(\frac{3}{4})^2.

    A. Find an expression for the height which the ball rises
    Problem. after it hits the floor for the nth time.

    \sum_{n=0}^{\infty} 10\cdot\frac{3}{4}^n

    Thus, H_0 = 10, H_1 = 10\frac{3}{4}, H_2 = 10(\frac{3}{4})^2 H_n = 10(\frac{3}{4})^n

    B. Find an expression for the total vertical distance the ball has travled for the first, second, third, and fourth times.

    D_n=H_0+2H_1+2H_3.......+2H_{n-1}

    D_1=10+2\cdot10(\frac{3}{4})


    D_2=10+2\cdot10(\frac{3}{4})+2\cdot10(\frac{3}{4})  ^2

    D_3=10+2\cdot10(\frac{3}{4})+2\cdot10(\frac{3}{4})  ^2+2\cdot10(\frac{3}{4})^3

    D_3=10+2\cdot10(\frac{3}{4})+2\cdot10(\frac{3}{4})  ^2+2\cdot10(\frac{3}{4})^3+2\cdot10(\frac{3}{4})^4

    D. Find the closed form expression for the total distance traveled for the nth time, this the part where I get my confused in properly making a simplify formula and simplifying it
    Last edited by RockHard; November 23rd 2009 at 05:09 PM.
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  2. #2
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    Why sum from 0 to infinity? Wait, actually, why sum at all?!

    after it's first dropped, the ball will rise to a height of 10 \cdot \frac{3}{4}. The second time, it will rise to a height of 10 \cdot \left( \frac{3}{4} \right)^2, and so on..

    So the nth rise is the nth term of the geometric series with a_0 = 10, \ q = \frac{3}{4}, which is..?
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  3. #3
    MHF Contributor
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    HI

    10+2(7.5)+2(5.625)+...

    Do you see why it must be multiplied by 2 ? Because we have to consider both the rebound and the path taken to fall again .

    10+2[7.5+5.625+...]

    10+2\sum^{n}_{r=1}10\cdot (\frac{3}{4})^r

     <br />
10+20\sum^{n}_{r=1}(\frac{3}{4})^r<br />
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  4. #4
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    Quote Originally Posted by Defunkt View Post
    Why sum from 0 to infinity? Wait, actually, why sum at all?!

    after it's first dropped, the ball will rise to a height of 10 \cdot \frac{3}{4}. The second time, it will rise to a height of 10 \cdot \left( \frac{3}{4} \right)^2, and so on..

    So the nth rise is the nth term of the geometric series with a_0 = 10, \ q = \frac{3}{4}, which is..?
    Defunkt, I was asking the same thing till I recongized, they are asking for total distance, when H_0 = 10 is the measurement the ball had to drop to the floor that what got my constants messed up when working with the closed form

    Thanks mathaddict I was looking for the correct constants to pull out, Ill try re-working the problem with your advice
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  5. #5
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    Ehhh, this is a tricky little son of a gun, they pulled the first H_1 out and up the summation as such

    10+20(\frac{3}{4})\sum_{n=1}^n(\frac{3}{4})^{n-1}

    However, I was able to match for a power of n
    Last edited by RockHard; November 23rd 2009 at 07:28 PM.
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