Answer
In order for a sequence to be geometric, the quotient of all consecutive terms must be constant.
Hence here: $\frac{a_{n+1}}{a_n}=\dfrac{3^{2(n+1)}}{3^{2n}}=3^{2}$, thus it is a geometric sequence.
$a_1=9$
$a_2=81$
$a_3=729$
$a_4=6561$
Work Step by Step
In order for a sequence to be geometric, the quotient of all consecutive terms must be constant.
Hence here: $\frac{a_{n+1}}{a_n}=\dfrac{3^{2(n+1)}}{3^{2n}}=3^{2}$, thus it is a geometric sequence.
$a_1=3^{2(1)}=9$
$a_2=3^{2(2)}=81$
$a_3=3^{2(3)}=729$
$a_4=3^{2(4)}=6561$
