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anonymous

  • 5 years ago

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  1. anonymous
    • 5 years ago
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    \[a_{1} = \sqrt{2} a _{2} = \sqrt{2\sqrt{2}} a3 = \sqrt{2\sqrt{2\sqrt{2}}} \] and so on... Find a recurrence relation for an+1. and find the limit

  2. anonymous
    • 5 years ago
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    this is what i get for 1st part ..\[a _{n+1} = \sqrt{2a _{n}}\]but the limits is??

  3. anonymous
    • 5 years ago
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    In your first example, is a_2 inside the radical?

  4. anonymous
    • 5 years ago
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    no... sorry for that ...i am new here...i keep space bar them but they still keep stick together = ="

  5. anonymous
    • 5 years ago
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    at 1st i guessing the limits to be square root 2... but that is stupid... i keep using calculator and i think the limit is 2..but i dunno how to show it

  6. anonymous
    • 5 years ago
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    Well, I think we can try to demonstrate the series in a general form and then it might make it easier to evaluate the limit. How's this: we know that \[\sqrt{2{\sqrt{2{\sqrt{2}}}}} = 2^{7/8}.\] Now? :)

  7. anonymous
    • 5 years ago
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    how to make this in nth terms .

  8. anonymous
    • 5 years ago
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    as Quantum said, we can rewrite the expression in a simpler way as: \[a_1=2^{1/2}a_2=2^{3/4}a_3=2^{7/8}a_4+...\] we can then easily see that: \[a_(n)=2^{{2n-1 \over 2n}}a_(n+1)\] now just rearrange to get a(n+1)

  9. anonymous
    • 5 years ago
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    ic.. thannks ...

  10. anonymous
    • 5 years ago
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    the final expression is:

  11. anonymous
    • 5 years ago
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    \[a_(n+1)=2^{{2n \over 2n-1}}a_n\] does that make sense?

  12. anonymous
    • 5 years ago
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    i getting \[a(n+1) = 2^{(2^{n}-1)/2^{n}}\]

  13. anonymous
    • 5 years ago
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    srry that is for a(n)

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