Question

Show that each sequence is geometric. Then find the common ratio and write out the first four terms $\left\{c_n\right\}=\left\{\frac{2^{n-1}}{4}\right\}$.

Short answer

The common ratio is $2$and the first four terms are $\frac{1}{4},\frac{1}{2},1,2$.

Step-by-step solution

Step 1. Write the given information.

We have given sequence.

$\left\{c_n\right\}=\left\{\frac{2^{n-1}}{4}\right\}$

The nth and (n-1)th term are

$$\begin{gathered} c_n=\frac{2^{n-1}}{4} \\ {c}_{n-1}=\frac{2^{(n-1)-1}}{4} \\ {c}_{n-1}=\frac{2^{n-2}}{4} \end{gathered}$$

Step 2. To find the common ratio. 

We have,

$$\begin{gathered} r=\frac{c_n}{c_{n-1}} \\ r=\frac{{\displaystyle \frac{2^{n-1}}{4}}}{{\displaystyle \frac{2^{n-2}}{4}}} \\ r=2^{n-1-n+2} \\ r=2 \end{gathered}$$

Since the ratio of the successive term is the non zero constant $2$, therefore $\left\{c_n\right\}$is a geometric sequence with common ratio is $2$.

Step 3. To find the terms. 

We have,

$$\begin{gathered} c_1=\frac{2^{1-1}}{4} \\ {c}_1=\frac{1}{4} \\ {c}_2=\frac{2^{2-1}}{4} \\ {c}_2=\frac{2}{4} \\ {c}_2=\frac{1}{2} \\ {c}_3=\frac{2^{3-1}}{4} \\ {c}_3=\frac{4}{4} \\ {c}_3=1 \\ {c}_4=\frac{2^{4-1}}{4} \\ {c}_4=\frac{8}{4} \\ {c}_4=2 \end{gathered}$$

Thus, the first four terms are$\frac{1}{4},\frac{1}{2},1,2$