Finding nth term of a sequence (explicit formula)

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Homework Help Overview

The discussion revolves around finding the nth term of a sequence defined by nested square roots involving the number 2. Participants are attempting to derive an explicit formula for the sequence.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster expresses difficulty in formulating the sequence's rule. One participant presents a potential general rule but notes it fails for the first term, prompting a request for assistance in refining it. Another participant suggests a method of expressing the terms using fractional exponents and hints at deriving a general term through induction.

Discussion Status

The conversation is active, with participants exploring different representations of the sequence and questioning the validity of proposed formulas. There is no explicit consensus, but suggestions for further exploration and refinement of the general term are being discussed.

Contextual Notes

Participants are working under the constraints of deriving a formula that accurately represents all terms of the sequence, including the first term, which has proven challenging.

demonelite123
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√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.
 
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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?
 
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}[/itex]
 

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