Question

Show that each sequence is geometric. Then find the common ratio and write out the first four terms$\left(u_n\right)=\left\{\frac{2^n}{3^{n-1}}\right\}$.

Short answer

The common ratio is$\frac{2}{3}$ and the first four terms are$2,\frac{4}{3},\frac{8}{9},\frac{16}{27}$.

Step-by-step solution

Step 1. Given information.  

We have given sequence $\left(u_n\right)=\left\{\frac{2^n}{3^{n-1}}\right\}$

The $nth$and $(n-1)th$terms are

$$\begin{gathered} u_n=\frac{2^n}{3^{n-1}} \\ {u}_{n-1}=\frac{2^{n-1}}{3^{(n-1)-1}} \end{gathered}$$

Step 2. To find the common ratio.   

We have,

$$\begin{gathered} r=\frac{u_n}{u_{n-1}} \\ r=\frac{{\displaystyle \frac{2^n}{3^{n-1}}}}{{\displaystyle \frac{2^{n-1}}{3^{n-2}}}} \\ r=\frac{2^{n-n+1}}{3^{n-1-n+2}} \\ r=\frac{2}{3} \end{gathered}$$

Since the ratio of the successive term is the non zero constant$\frac{2}{3}$therefore,$\left(u_n\right)$ is a geometric sequence with common ratio is$\frac{2}{3}$.

Step 3. To find the terms.   

We have,

$$\begin{gathered} u_1=\frac{2^1}{3^{1-1}} \\ {u}_1=2 \\ {u}_2=\frac{2^2}{3^{2-1}} \\ {u}_2=\frac{4}{3} \\ {u}_3=\frac{2^3}{3^{3-1}} \\ {u}_3=\frac{8}{9} \\ {u}_4=\frac{2^4}{3^{4-1}} \\ {u}_4=\frac{16}{27} \end{gathered}$$

Thus, the first four terms are$2,\frac{4}{3},\frac{8}{9},\frac{16}{27}$.