## College Algebra (10th Edition)

$a_n=\dfrac{1}{2^{n-1}}$
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: $a_n=\dfrac{1}{2^{n-1}}$