## College Algebra (10th Edition)

$a_n=\left(\dfrac{2}{3}\right)^n$
The pattern shows that: (1) The numerators involve powers of $2$. first term's numerator = $2^1=2$ second term's numerator = $2^2=4$ third term's numerator = $2^3=8$ fourth term's numerator = $2^4=16$ (2) The denominators involve powers of $3$: first term's denominator = $3^1 = 3$ second term's denominator = $3^2 =9$ third term's denominator = $3^3 = 27$ fourth term's denominator = $3^4=81$ Thus, the $n^{th}$ term of the sequence is: $a_n=\dfrac{2^n}{3^n}=\left(\dfrac{2}{3}\right)^n$