$\{ x \in \mathbb{C}$ | $x^{2n+1} = 1$ , $x \neq 1\}$
Compute $\displaystyle{\sum_{i=1}^{2n} \frac{x^{2i}}{x^i-1}}$
$\{ x \in \mathbb{C}$ | $x^{2n+1} = 1$ , $x \neq 1\}$
Compute $\displaystyle{\sum_{i=1}^{2n} \frac{x^{2i}}{x^i-1}}$
Since
$(x^{\small{2n+1}}-1) = (x-1)(x^{\small{2n}}+x^{\small{2n-1}}+\cdots+x^{\small{2}}+x+1)=0$
Therefore $(x^{\small{2n}}+x^{\small{2n-1}}+\cdots+x^{\small{2}}+x+1)=0 \hspace{4pt} ({\text{since}} \hspace{4pt} x \neq 1) \tag{1}$
$ \begin{align*} S = \sum_{i=1}^{\small{2n}} \frac{x^{\small{2i}}}{x^i-1} &= \sum_{i=1}^{\small{2n}} x^i + \sum_{i=1}^{\small{2n}} \frac{x^{\small{i}}}{x^i-1}\\ &= -1 + \sum_{i=1}^{\small{2n}} \frac{x^{\small{i}}}{x^i-1} \hspace{12pt} {\text{from}} \hspace{4pt} (1) \\ &= -1 + \sum_{i=1}^{n} \frac{x^{\small{i}}}{x^i-1} + \sum_{i={\small{n+1}}}^{\small{2n}} \frac{x^{\small{i}}}{x^i-1} \tag{2} \end{align*} $
Since $x^{\small{2n+1}} = 1$
$ \frac{x^{\small{n+i}}}{x^{\small{n+i}}-1} = \frac{-1}{x^{\small{n+1-i}}-1} $
Therefore $(2)$ can be simplified to
$ \begin{align*} & -1 + \sum_{i=1}^{n} \frac{x^{\small{i}}}{x^i-1} + \sum_{i={\small{n}}}^{\small{1}} \frac{-1}{x^{\small{n+1-i}}-1}\\ &= -1 + \sum_{i=1}^{n} \frac{x^{\small{i}}}{x^i-1} + \sum_{i={\small{1}}}^{\small{n}} \frac{-1}{x^{\small{i}}-1}\\ &= -1 + \sum_{i=1}^{n} \frac{x^{\small{i}}-1}{x^i-1} \\ &= n-1 \end{align*} $