An equality of two finite sums

Let $n \in \mathbb{N}$. Prove that:

$$\sum_{k=1}^{n}\frac{1}{4k^{2}-2k}=\sum_{k=n+1}^{2n}\frac{1}{k}$$

Solution

We have successively that:

\begin{align*}
\sum_{k=1}^{n}\frac{1}{4k^2-4k} &=\sum_{k=1}^{n}\frac{1}{2k(2k-1)} \\
 &=\frac{1}{2}\sum_{k=1}^{n} \left [ \frac{2}{2k-1} - \frac{1}{k} \right ] \\
 &=\sum_{k=1}^{n} \frac{1}{2k-1} - \sum_{k=1}^{n} \frac{1}{2k} \\
 &\overset{(*)}{=}\sum_{k=1}^{2n} \frac{1}{k} - \sum_{k=1}^{n}\frac{1}{2k} - \sum_{k=1}^{n} \frac{1}{2k} \\
 &= \sum_{k=1}^{2n} \frac{1}{k}  - \sum_{k=1}^{n} \frac{1}{k} \\
 &= \sum_{k=n+1}^{2n} \frac{1}{k}
\end{align*}

$(*)$ since it holds that:

$$\sum_{k=1}^{2n} \frac{1}{k} = \sum_{k=1}^{n} \frac{1}{2k-1}  - \sum_{k=1}^{n}  \frac{1}{2k} $$