Question

In Problems 29–36, the given pattern continues. Write down the nth term of a sequence $\left\{a_n\right\}$suggested by the pattern.

$1,\frac{1}{2},\frac{1}{4},\frac{1}{8},...$

Short answer

The nth term of the given sequence is$a_n=\frac{1}{2^{n-1}}$.

Step-by-step solution

Step 1. Write the given information.

The given sequence is:

$1,\frac{1}{2},\frac{1}{4},\frac{1}{8},...$

Step 2. Determine the pattern of the sequence and write down the nth term. 

$a_1=1,a_2=\frac{1}{2},a_3=\frac{1}{4},a_4=\frac{1}{8}$

The pattern of the sequence is:

$$\begin{gathered} a_1=\frac{1}{2^1-1}=1 \\ {a}_2=\frac{1}{2^{2-1}}=\frac{1}{2} \\ {a}_3=\frac{1}{2^{3-1}}=\frac{1}{4} \\ {a}_4=\frac{1}{2^{4-1}}=\frac{1}{8} \\ {a}_n=\frac{1}{2^{n-1}} \end{gathered}$$