Let N be an even integer, and $x[n]$ be a finite length signal over the interval $n\in [0,N-1]$; let $X[k]$ be the N-pt DFT of $x[n]$.

Find the 2N-Pt DFT of $x_1[n]$ in terms of $x[n]$, where:

$$x_1[n] = \begin{cases} x[\frac{n}{2}], & \text{if } n \text{ is even}\\ 0 & \text{else} \end{cases} $$

Now, the idea I had in mind was to write out the DFT sum and then change the summing variable $n\rightarrow 2n$ but that doesn't work:

$$X_1(k) = \sum_{n=0}^{2N-1} x[n/2]e^{-j\frac{2\pi}{2N}nk} $$

$$\downarrow n \text{ to } 2n$$

$$X_1(k) = \sum_{n=0}^{N-1/2} x[n]e^{-j\frac{2\pi}{N}nk} \neq X(k)$$

This doesn't work for two reasons, 1) The form of the DFT is not correct and 2) the fact that $x_1[n] = x[n/2]$ for only even $n$'s is not reflected in the sum!

Any help would be appreciated!