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Finding nth term of a sequence (explicit formula)

  1. Feb 12, 2009 #1
    √2, √(2√2), √(2√(2√2) ), √(2√(2√(2√2) ) )

    this sequence has been giving me a lot of trouble. i have no idea how to write the fomula. please help me.
     
  2. jcsd
  3. Feb 12, 2009 #2
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  4. Feb 12, 2009 #3
    well i did this so far:
    √2, √2^(3/2), √2^(5/2), √2^(7/2)

    and i got the general rule:
    √2^(2n-1 / 2)

    the only problem is that it works for everything but the first term. can someone please help me fix up my rule so that it fits all of the terms?
     
  5. Feb 13, 2009 #4

    HallsofIvy

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    No that doesn't work for any term, but writing them in terms of fractional exponents is the right way to go. You have [itex]\sqrt{2}= 2^{1/2}[/itex], [itex]\sqrt{2\sqrt}{2}}= (2(2^{1/2})^{1/2}= (2^{3/2})^{1/2}= 2^{3/4}[/itex], etc., [itex]\sqrt{2\sqrt{2\sqrt}{2}}}= (2(2^{3/4})^{1/2}=(2^{7/4})^{1/2}= 2^{7/8}[/itex], etc..

    You should be able to guess the general term from that. And, in fact, you should be able to prove it using induction and the fact that [itex]a_{n+ 1}= \sqrt{2a_n}
     
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