Expressing 2N point DFT in terms of N point DFT I have a problem with expressing odd samples of X2 in terms of X1. I understand that the resulting DFT will be more precise in terms of expressing the exact spectrum of signal x[n], due to more samples. Moreover I know that even samples of X2(k) are copy of the spectrum of X1(k), however I do not know how to mathematically compute the odd ones.

The task is as follows:

Let $x[n]$  be  a  periodic  sequence  with  period $N_1$ .   Thus $x[n]$  is  also  periodic  for  period $N_2 =  2 N_1$ .   We  may compute $X_1[k]$:  $N$-point DFT of $x[n]$ and $X_2[k]$:  $2N$-point DFT of $x[n]$. 


Express $X_2$in terms of $X_1$ Hint: it is easy with even samples $X_2[2m]$ , harder with odd ones $X2_[2m+ 1]$ , for $m$ integer.
Consider a sequence $x[n]$ of length $N$ whose $N$-point DFT is $X[k]$. Then let $X_2[k]$ be the $2N$-point DFT of $x[n]$. 

As you have stated, the even indexed samples of $X_2[k]$ will be easily shown to be: $$X_2[k] = X[k/2] ~~~,~~~k = 2m, m=0,1,...,N-1$$

Then the odd indexed samples of $X_2[k]$ will be given by the $N$-point DFT of the signal  $x[n] e^{-j \frac{\pi}{N} n } $, $n=0,1,...,N-1$. 

$$X_{2}[k] = \sum_{n=0}^{N-1} x[n] e^{-j \frac{\pi}{N} n} e^{-j \frac{2\pi}{N} k n}  = X[k + 0.5]  ~~~~ ,~~~ k =2m+1, m=0,1,...,N-1 $$