Answer
$a_{n}=2^{(2^{n}-1)/2^{n}}$
Work Step by Step
$k=1, a_{1}=2^{1/2}$
$k=2, a_{2}=(2\cdot 2^{1/2})^{1/2}=(2^{3/2})^{1/2}=2^{3/4}$
$k=3, a_{3}=(2a_{2}^{1/2})^{1/2}=(2\cdot 2^{3/4})^{1/2}=(2^{7/4})^{1/2}=2^{7/8}$
$k=4, a_{3}=(2a_{3}^{1/2})^{1/2}=(2\cdot 2^{7/8})^{1/2}=(2^{15/4})^{1/2}=2^{15/16}$
The pattern that is emerging for the exponent is
k=1, $ \ \ 1=2^{1}-1, \ \ 2=2^{1}$
k=2, $ \ \ 3=2^{2}-1, \ \ 4=2^{2}$
k=3, $ \ \ 7=2^{3}-1, \ \ 8=2^{3}$
k=4, $ \ \ 15=2^{4}-1, \ \ 16=2^{4}$
So, for $a_{n}, $
n-th numerator=$2^{n}-1$
n-th denominator$= 2^{n}$
$a_{n}=2^{(2^{n}-1)/2^{n}}$
