IDFT of upsampling in frequency domain
This answer does not require any elaborate mathematics. Rather it requires a careful interperation of periodicity of DFT sequences considering their lengths.

Lets display the signal chain below:

$$ x[n] \longrightarrow \boxed{ 32DFT} \longrightarrow  X[k] \longrightarrow \boxed{ \uparrow 2 } \longrightarrow Y[k] \longrightarrow \boxed{ 64-IDFT } \longrightarrow  y[n] $$

Now the signals are:

Time sequence $x[n]$ has a length of $32$ samples defined for $n=0,1,...,31$. For any other values of $n$, we interpret $x[n]$ as the periodic extension by the help of modulus operator as $x[n] = x[(n)_{32}]$. For example $x[35] = x[ (35)_{32} ] = x[3]$.  

The frequency sequence $X[k]$ has a length of $32$ samples defined for $k=0,1,...,31$. For any other values of $k$, we interpret $X[k]$ as the $32$ point periodic extension of $X[k]$ similarly.

After the expansion block we have:

DFT frequency sequence $Y[k]$ has a length of $64$ samples defined for $k=0,1,...,63$. For any other values of $k$, we interpret $Y[k]$ as the periodic extension by $64$ samples similarly.

The IDFT signal $y[n]$ has a length of $64$ samples defined for $n=0,1,...,63$. For any other values of $n$, we interpret $y[n]$ as the periodic extension by $64$ samples similarly.

Now, simply write the $64$ point inverse DFT equation to define $y[n]$ :

$$
\begin{align}
y_{64}[n] &amp;= \frac{1}{64} \sum_{k=0}^{63} Y[k] e^{-j \frac{2\pi}{64} k n } ~~~&amp;,~~~ \text{ for } k,n = 0,1,...,63 \\
&amp;= \frac{1}{64} \sum_{k=0}^{63} X[k/2] e^{-j \frac{2\pi}{64} k n } ~~~&amp;,~~~ \text{ for } k,n = 0,1,...,63 \\
&amp;= \frac{1}{64} \sum_{m=0, k=2m}^{31} X[m] e^{-j \frac{2\pi}{64} 2m n } ~&amp;,~ \text{ for } m=0,1,..,31 ~~ n = 0,1,...,63 \\
&amp;= \frac{1}{2} \left( \frac{1}{32} \sum_{m=0}^{31} X[m] e^{-j \frac{2\pi}{32} m n } \right) ~~&amp;,~\text{ for } m=0,1,..,31 ~~~ n = 0,1,...,63 \\
y_{64}[n] &amp;= \frac{1}{2} x[(n)_{32}] ~~~&amp;,~~~ \text{ for } n = 0,1,...,63 \\
\end{align}
$$

The expression in the parenthesis above is simply the $32$ point inverse DFT of the sequence $x[n]$. Note that this sequence has a length of $32$ points and we can express it with the modulus notation $x[n] = x[(n)_{32}]$ as in the last line.

And the last line describes the $64$ point signal $y_{64}[n]$ as a two times repetition of $32$ point $x[n]$, divided by $2$. In other words, $y[0] = x[0], y[15]=x[15], y[40] = x[8], y[60]= x[28]$ etc...