Answer
$a_1=\sqrt{3}=3^{1/2}$
$a_2=3^{3/4}$
$a_{3}=3^{7/8}$
$a_{4}=3^{15/16}$
$a_{5}=3^{31/32}$
$a_6=3^{63/64}$
$a_7=3^{127/128}$
Work Step by Step
We are given:
$a_{n}=\sqrt{3a_{n-1}}$ and $a_1=\sqrt{3}$
We evaluate:
$a_2=\sqrt{3a_1}=\sqrt{3\sqrt{3}}=(3*3^{1/2})^{1/2}=3^{3/4}$
$a_{3}=\sqrt{3a_{2}}=\sqrt{3^1*3^{3/4}}=\sqrt{3^{7/4}}=3^{7/8}$
$a_{4}=\sqrt{3a_{3}}=\sqrt{3^1*3^{7/8}}=\sqrt{3^{15/8}}=3^{15/16}$
$a_{5}=\sqrt{3a_4}=\sqrt{3^1*3^{15/16}}=\sqrt{3^{31/16}}=3^{31/32}$
$a_6=\sqrt{3a_{5}}=\sqrt{3^1*3^{31/32}}=\sqrt{3^{63/32}}=3^{63/64}$
$a_7=\sqrt{3a_6}=\sqrt{3^1*3^{63/64}}=\sqrt{3^{127/64}}=3^{127/128}$
