Answer
$a_n=\dfrac{1}{2^{n-1}}$
Work Step by Step
The pattern shows that:
(1) The numerator is always 1.
(2) The denominators involve the powers of $2$:
first term's denominator = $2^{1-1} = 2^0=1$
second term's denominator = $2^{2-1} = 2^1=2$
third term's denominator = $2^{3-1} = 2^2=4$
fourth term's denominator = $2^{4-1}=2^3=8$
Notice the exponent of $2$ is equal to the term number minus 1 (or $n-1$).
Thus, the denominator is represented by $2^{n-1}$.
Therefore, the $n^{th}$ term of the sequence is:
$a_n=\dfrac{1}{2^{n-1}}$
