Question

consider this...

Answer 1

\[a_{1} = \sqrt{2} a _{2} = \sqrt{2\sqrt{2}} a3 = \sqrt{2\sqrt{2\sqrt{2}}} \]

and so on... Find a recurrence relation for an+1. and find the limit

Answer 2

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

Answer 3

In your first example, is a_2 inside the radical?

Answer 4

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

Answer 5

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

Answer 6

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? :)

Answer 7

how to make this in nth terms .

Answer 8

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)

Answer 9

ic.. thannks ...

Answer 10

the final expression is:

Answer 11

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

Answer 12

i getting \[a(n+1) = 2^{(2^{n}-1)/2^{n}}\]

Answer 13

srry that is for a(n)