## Precalculus: Mathematics for Calculus, 7th Edition

$a_{n}=2^{(2^{n}-1)/2^{n}}$
$k=1, a_{1}=2^{1/2}$ $k=2, a_{2}=(2\cdot 2^{1/2})^{1/2}=(2^{3/2})^{1/2}=2^{3/4}$ $k=3, a_{3}=(2a_{2}^{1/2})^{1/2}=(2\cdot 2^{3/4})^{1/2}=(2^{7/4})^{1/2}=2^{7/8}$ $k=4, a_{3}=(2a_{3}^{1/2})^{1/2}=(2\cdot 2^{7/8})^{1/2}=(2^{15/4})^{1/2}=2^{15/16}$ 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$= 2^{n}$ $a_{n}=2^{(2^{n}-1)/2^{n}}$