Finding nth term of a sequence (explicit formula)

[demonelite123]

√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.

 

[JG89]

Ignore Post

 

[demonelite123]

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?

 

[HallsofIvy]

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