## anonymous 5 years ago consider this...

1. anonymous

$a_{1} = \sqrt{2} a _{2} = \sqrt{2\sqrt{2}} a3 = \sqrt{2\sqrt{2\sqrt{2}}}$ and so on... Find a recurrence relation for an+1. and find the limit

2. anonymous

this is what i get for 1st part ..$a _{n+1} = \sqrt{2a _{n}}$but the limits is??

3. anonymous

4. anonymous

no... sorry for that ...i am new here...i keep space bar them but they still keep stick together = ="

5. anonymous

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

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

how to make this in nth terms .

8. anonymous

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

ic.. thannks ...

10. anonymous

the final expression is:

11. anonymous

$a_(n+1)=2^{{2n \over 2n-1}}a_n$ does that make sense?

12. anonymous

i getting $a(n+1) = 2^{(2^{n}-1)/2^{n}}$

13. anonymous

srry that is for a(n)