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Exhaustion methods

  1. Mar 13, 2013 #1
    Well if anyone has read Apostol's vol 1. Can anyone explain in the beginingthe sum of the areas of the outer rectangles is (b^3/n^3)(k^2) or
    (b^3/n^3)(1^2+2^2+3^2…+n^2)

    And the area of the inner rectangles and (b^3/n^3)(1^2+2^2+3^2…+(n-1)^2)

    What I would like to know is why does the value of k for the inner area sum goes to n-1
    Thank you
     
  2. jcsd
  3. Mar 13, 2013 #2
    Oh yeah. My idea was that the interior rectangles cannot have a value directly at b so the value of k must be under the of n so n doesn't cancek out. Although this woukd not occur anyways because the value of n on the bottom is n^3 and the value of k on tome is only squared. Honestly though I don't have a good idea. I tryed it graphically on my own paper I got lost with what I was doing.
     
  4. Mar 14, 2013 #3

    HallsofIvy

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    You should know that
    [tex]1^2+ 2^2+ 3^2+ \cdot\cdot\cdot+ n^2= \frac{n(n+1)(2n+1)}{6}[/tex]

    As to what happens to the "k", I doubt if anyone can answer because you have not said what "k" represents.
     
  5. Mar 14, 2013 #4
    Nvm I got it. Thanks though
     
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