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Math Help - Problem with Area and Sequences/Series

  1. #1
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    Problem with Area and Sequences/Series

    Hi everyone! I was having this problem in textbook homework regarding sequences and series, but having to do with area which threw me off. There is a picture necessary so I have a picture of the question and picture to go with it. Thanks for all the help!

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  2. #2
    Junior Member Spimon's Avatar
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    Does the text book have an answer? Below is my solution. I hope it's right!



    This would mean after 9 repetitions, we would have 127.75 sq.in.

    I hope thats right.
    Last edited by Spimon; September 21st 2007 at 01:07 AM.
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  3. #3
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    I think you need to figure a \sqrt{2} into your series.

    The sides of the triangles of the successive iterations are \frac{1}{2} the hypotenuse of the earlier triangles forming the sides of the inner squares.

    -Scott
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by ScottO View Post
    I think you need to figure a \sqrt{2} into your series.

    The sides of the triangles of the successive iterations are \frac{1}{2} the hypotenuse of the earlier triangles forming the sides of the inner squares.

    -Scott
    Succesive squares have side 1/sqrt(2) that of the preceeding square, so the
    areas of corresponding similar sub areas decrease by a factor of 2 at each step.

    RonL
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  5. #5
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    I aways seem to have a knack for finding a route that's a bit more complicated.

    My thinking was the sides of the squares were \frac{s}{\sqrt{2}}, so their areas were 1/2, and the areas of pairs of shaded triangles were 1/4.

    So I put down this series...

    \frac{1}{4}\sum(\frac{s}{(\sqrt{2})^{n-1}})^2

    Which I see now, can be simplified to \sum\frac{s^2}{2^{n+1}}, the same thing Spimon shows above.

    -Scott
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