Question

Show that each sequence is geometric. Then find the common ratio and write out the first four terms$\left(e_n\right)=\left(2^{n/3}\right)$

Short answer

The common ratio is $2^{1/3}$and the first four terms are $2^{1/3},2^{2/3},2,2^{4/3}$.

Step-by-step solution

Step 1. Given information.  

We have given sequence $\left(e_n\right)=\left\{2^{n/3}\right\}$.

The localid="1646927031642" $nth$and $(n-1)$th terms are

$$\begin{gathered} e_n=2^{n/3} \\ {e}_{n-1}=2^{n-1/3} \end{gathered}$$

Step 2. To find the common ratio.   

We have,

$$\begin{gathered} r=\frac{e_n}{e_{n-1}} \\ r=\frac{2^{n/3}}{2^{n-1/3}} \\ r=2^{\frac{n-n+1}{3}} \\ r=2^{\frac{1}{3}} \end{gathered}$$

Since the ratio of the successive term is the non zero constant $2^{\frac{1}{3}}$

therefore,$\left(e_n\right)$is a geometric sequence with common ratio is$2^{\frac{1}{3}}$.

Step 3. To find the terms.  

We have,

$$\begin{gathered} e_1=2^{\frac{1}{3}} \\ {e}_2=2^{\frac{2}{3}} \\ {e}_3=2^{\frac{3}{3}} \\ {e}_3=2 \\ {e}_4=2^{\frac{4}{3}} \end{gathered}$$

Thus, the first four terms are$2^{\frac{1}{3}},2^{\frac{2}{3}},2,2^{\frac{4}{3}}$.