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A difficult one

  1. Dec 2, 2005 #1
    express (2^1/2-1)^10 in the form k^1/2-(k-1)^1/2 where k is a positive integer.{the square roots need not be irrational}we can do this by binomial theorem but it is very tedious.is there a short & appropriate method to solve this problem?
     
  2. jcsd
  3. Dec 3, 2005 #2

    Tide

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    Just solve the equation

    [tex]\sqrt{k} - \sqrt{k-1} = (\sqrt 2 - 1)^{10}[/tex]

    algebraically for k. I get k = 11,309,769. This will be a mess unless you try something like

    [tex]\sqrt k - \sqrt {k-1} = N[/tex]

    from which

    [tex]\sqrt k + \sqrt {k-1} = \frac {1}{N}[/tex]

    leading to

    [tex]2\sqrt k = N + \frac {1}{N}[/tex]

    This is easy to solve for k and the solution can be simplified to what I showed above.
     
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