Answer
The sequence is geometric with a common ratio of $2$.
The first four terms are:
$c_1=\frac{1}{4}$
$c_2=\frac{1}{2}$
$c_3=1$
$c_4=2$
Work Step by Step
Substitute $1, 2, 3,$ and $4$ for $n$ into the given formula:
$c_1=\dfrac{2^{1-1}}{4} = \dfrac{2^0}{4}=\dfrac{1}{4}$
$c_2=\dfrac{2^{2-1}}{4} = \dfrac{2^1}{4}=\dfrac{2}{4}=\dfrac{1}{2}$
$c_3=\dfrac{2^{3-1}}{4} = \dfrac{2^2}{4}=\dfrac{4}{4}=1$
$c_4=\dfrac{2^{4-1}}{4} = \dfrac{2^3}{4}=\dfrac{8}{4}=2$
Notice the the next term is equal to $2$ times the current term.
This means that a common ratio of $2$ exists, and the sequence is geometric.