Chapter 11 - Sequences; Induction; the Binomial Theorem - Section 11.1 Sequences - 11.1 Assess Your Understanding - Page 827: 29

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

We notice in the sequence pattern that the numerator is constant at $1$, whereas the denominator starts from $1$ and gets multiplied by $2$ each time. So, we determine the pattern as: $\{a_n\}=\dfrac{1}{2^{n-1}}$ . Thus, $a_1=\dfrac{1}{2^{1-1}}=\dfrac{1}{2^0}=1$ and $a_2=\dfrac{1}{2^{2-1}}=\dfrac{1}{2^1}=\frac{1}{2}$, etc.