Discrete Fourier Series of square of a signal I am having trouble in deriving a result, which asks me to find a relation between $X_{1}(k)$ and $X_{2}(k)$ where $X_{1}(k)$ are the DFS coefficients of single period of signal $x(n)$ and $X_{2}(k)$ are the DFS coefficients of two periods of the same $x(n)$. My approach is this.

$$X_{1}(k) = \sum_{n=0}^{N-1}x(n)W_{N}^{-kn}$$
and $$X_{2}(k) = \sum_{n=0}^{2N-1}x(n)W_{2N}^{-kn}$$
Since we know that $W_{N} = W_{2N}^{2}$. That means that, if I square $X_{2}(k)$, then I get this
$$X_{2}^{2}(k)=\sum_{n=0}^{2N-1}x^{2}(n)W_{2N}^{-2kn}=\sum_{n=0}^{2N-1}x^{2}(n)W_{N}^{-kn}$$
Now this is where I can't move further. What should I do with the last expression. I mean what are the DFS coefficients of the square of a sequence? And also note that now we are over two periods ($2N-1$).
Here is an outline (in case it's homework)


Use a different frequency index for $X_2(l)$
Split the $X_2(l)$ into two sums: from $0:N-1$ and from $N-1:2N$
Use the periodicity of x[n], i.e.  $x(n) = x(n+N)$
Substitute $m = n-N$ in the second sum
Recombine the two sums again
Stare what happens to even $l = 2k$ and odd $l = 2k+1$ terms