If the Z-transform of a finite-duration discrete-time signal $x[n]$ is $X(z)$, then the Z transform of the signal $y[n]=x[2n]$ is

The correct option is:
$$Y(z)=\frac{1}{2}[X\left(z^{1/2}\right)+X(-z^{1/2})]$$

Step-by-Step Explanation:

1. Definition of Z-transform:
The Z-transform of a discrete-time signal $x[n]$ is defined as:
$$X(z)=\sum _{n=-\infty }^{\infty }x[n]z^{-n}$$

2. Understanding Decimation (Downsampling by 2):
The given signal is $y[n]=x[2n]$. This is the downsampled version of $x[n]$ by a factor of 2, meaning it consists only of the even-indexed samples of $x[n]$.

To represent $y[n]$ in terms of $x[n]$, we can define an intermediate signal $x_e[n]$ that keeps the even-indexed samples of $x[n]$ and sets the odd-indexed samples to zero:
$$x_e[n]=x[n]\cdot \frac{1+{(-1)}^n}{2}$$ When $n$ is even, ${(-1)}^n=1$, so $x_e[n]=x[n]$. When $n$ is odd, ${(-1)}^n=-1$, so $x_e[n]=0$.

3. Z-transform of the Even Sequence:
We compute the Z-transform of $x_e[n]$:
$$X_e(z)=\sum _{n=-\infty }^{\infty }{x}_e[n]z^{-n}=\sum _{n=-\infty }^{\infty }x[n]\left(\frac{1+{(-1)}^n}{2}\right)z^{-n}$$ Splitting this sum into two components gives:
$$X_e(z)=\frac{1}{2}\sum _{n=-\infty }^{\infty }x[n]z^{-n}+\frac{1}{2}$$ Using the definition of the Z-transform, we get:
$$X_e(z)=\frac{1}{2}[X(z)+X(-z)]$$

4. Relating $y[n]$ to $x_e[n]$:
Since $y[n]=x[2n]$, we have $x_e[2n]=y[n]$ and $x_e[n]=0$ for all odd $n$. Therefore, we can write:
$$X_e(z)=\sum _{n=-\infty }^{\infty }{x}_e[n]z^{-n}=\sum _{k=-\infty }^{\infty }{x}_e[2k]z^{-2k}=\sum _{k=-\infty }^{\infty }y[k]{(z^2)}^{-k}=Y(z^2)$$ Comparing the two expressions for $X_e(z)$:
$$Y(z^2)=\frac{1}{2}[X(z)+X(-z)]$$

5. Substituting $z$ with $z^{1/2}$:
To solve for $Y(z)$, we replace $z$ with $z^{1/2}$ in the equation above:
$$Y(z)=\frac{1}{2}[X\left(z^{1/2}\right)+X(-z^{1/2})]$$ This perfectly matches the correct option.