Answer
$2^{1/2},2^{-1/4},2^{-5/8},2^{-13/16},2^{-29/32}$
Work Step by Step
We are given the sequence:
$a_1=\sqrt 2$
$a_n=\sqrt{\dfrac{a_{n-1}}{2}}$
Determine the first 5 terms of the sequence by substituting 1, 2, 3, 4, 5 for $n$:
$a_1=\sqrt 2$
$a_2=\sqrt{\dfrac{a_1}{2}}=\sqrt{\dfrac{\sqrt 2}{2}}=\sqrt {2^{-1/2}}=2^{-1/4}$
$a_3=\sqrt{\dfrac{a_2}{2}}=\sqrt{\dfrac{2^{-1/4}}{2}}=\sqrt{2^{-5/4}}=2^{-5/8}$
$a_4=\sqrt{\dfrac{a_3}{2}}=\sqrt{\dfrac{2^{-5/8}}{2}}=\sqrt{2^{-13/8}}=2^{-13/16}$
$a_5=\sqrt{\dfrac{a_4}{2}}=\sqrt{\dfrac{2^{-13/16}}{2}}=\sqrt{2^{-29/16}}=2^{-29/32}$
