The correct answer is 2 + 2e^{(j$\frac{2\pi }{3}$n)} cos($\frac{2\pi }{6}$n), which corresponds to the expression shown in the option images.

Step-by-Step Explanation:

1. Understand the Given Information:
From the image in the question statement:

We are given a discrete-time periodic signal $x\left[n\right]$ with period $N=3$.
The Discrete-Time Fourier Series (DTFS) coefficients are periodic with period $N=3$, meaning $a_{k+N}=a_k$ for all integer $k$.
The given non-zero Fourier series coefficients are:

$a_{-3}=2$

$a_4=1$

2. Find the Coefficients in the Fundamental Period (k = 0, 1, 2):
Since the DTFS coefficients $a_k$ are periodic with $N=3$:

$a_0=a_{-3+3}=a_{-3}=2$

$a_1=a_{4-3}=a_4=1$

$a_2=0$ (since no other non-zero coefficient is given in the period)

3. Reconstruct the Signal x[n]:
The synthesis formula for DTFS is:

$x\left[n\right]=\sum _{k=0}^{N-1}{a}_k{e}^{jk\left(\frac{2\pi }{N}\right)n}$

$x\left[n\right]=a_0+a_1{e}^{j\left(\frac{2\pi }{3}\right)n}=2+e^{j\left(\frac{2\pi }{3}\right)n}$

4. Express in Terms of Trigonometric Form:
Using Euler's identity $2\cos \left(\theta \right)=e^{j\theta }+e^{-j\theta }$, we note that for $\theta =\frac{2\pi }{6}n=\frac{\pi }{3}n$:

$2\cos \left(\frac{2\pi }{6}n\right)=e^{j\frac{\pi }{3}n}+e^{-j\frac{\pi }{3}n}$

$2e^{j\left(\frac{2\pi }{3}\right)n}\cos \left(\frac{2\pi }{6}n\right)=e^{j(\frac{2\pi }{3}+\frac{\pi }{3})n}+e^{\mathrm{j}(\frac{2\pi }{3}-\frac{\pi }{3})n}=e^{j\pi n}+e^{j\frac{\pi }{3}n}$

$x\left[n\right]=1+2e^{j\left(\frac{2\pi }{3}\right)n}\cos \left(\frac{2\pi }{6}n\right)$