## Precalculus (10th Edition)

$\{s_n\}=\frac{2^n}{3^n}=(\frac{2}{3})^n$.
The pattern suggests that $\{s_n\}=\frac{2^n}{3^n}=(\frac{2}{3})^n$. Because the numerator begins at $2$ and we can see that it gets multiplied by $2$ at each element, and the denominator begins at $3$ and we can see that it gets multiplied by $3$ at each element.