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 ($2N-1$).