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A problem: help needed

  1. Nov 16, 2006 #1
    this is the problem:
    let {ak} be a sequence of real numbers. and the sequence is defined as

    [tex]a_0 = 2006[/tex]

    [tex]a_{n} = \log_{2}a_{n-1}[/tex]

    now i have to find out the number of terms in the sequence.

    this is what i have done:
    i see that a5 is negative. so we can't find out a6 or any other subsequent terms. so the number of terms is 6 (a0 upto a5).

    but how can i find out the number of terms without calculating the terms upto a5? is there a way to do it?
     
  2. jcsd
  3. Nov 16, 2006 #2

    Office_Shredder

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    If [tex]a_{n} = \log_{2}a_{n-1}[/tex] is true, then [tex]2^{a_n} = a_{n-1)[/tex]

    So this tells us [tex]2^{2^{2...^{2^{a_n}}}} = a_0[/tex] So find how many times you can put two to itself before it goes past 2006.

    I think that works
     
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