Answer
$a_1= \sqrt2$
$a_2 =\dfrac{\sqrt{2\sqrt2}}{2} \\ a_3 =\dfrac{\sqrt[4]{2\sqrt2}}{2} \\ a_4 = \dfrac{\sqrt[8]{2\sqrt2}}{2} \\ a_5= \dfrac{\sqrt[16]{2\sqrt2}}{2}$
Work Step by Step
We are given that $a_n=\sqrt{\dfrac{a_n}{n}} $ and $a_1=\sqrt2$. In order to determine the remaining values, we will have to substitute $n=2,3,4,5$ into the given sequence. We use previous terms to compute the successive terms, as follows:
$a_1= \sqrt2$
$a_2 = \sqrt{\dfrac{a_1}{2}} = \sqrt{\dfrac{\sqrt2}{2}}=\sqrt{\dfrac{\sqrt2}{2} \times \dfrac{2}{2}}=\dfrac{\sqrt{2\sqrt2}}{2} \\ a_3 = \sqrt{\dfrac{a_2}{2}}=\sqrt{\dfrac{\frac{\sqrt{2\sqrt2}}{2}}{2}}=\sqrt{\dfrac{\sqrt{2\sqrt2}}{4}}=\dfrac{\sqrt{\sqrt{2\sqrt2}}}{2}=\dfrac{\sqrt[4]{2\sqrt2}}{2} \\ a_4 = \sqrt{\dfrac{a_3}{2}}=\sqrt{\dfrac{\frac{\sqrt[4]{2\sqrt2}}{2}}{2}}=\sqrt{\dfrac{\sqrt[4]{2\sqrt2}}{4}}=\dfrac{\sqrt{\sqrt[4]{2\sqrt2}}}{2}=\dfrac{\sqrt[8]{2\sqrt2}}{2} \\ a_5= \sqrt{\dfrac{a_4}{2}}=\sqrt{\dfrac{\frac{\sqrt[8]{2\sqrt2}}{2}}{2}}=\sqrt{\dfrac{\sqrt[8]{2\sqrt2}}{4}}=\dfrac{\sqrt{\sqrt[8]{2\sqrt2}}}{2}=\dfrac{\sqrt[16]{2\sqrt2}}{2}$
