<s>[INST] how can I prove this $\sum_{k=1}^{n}\frac{1}{4k^{2}-2k}=\sum_{k=n+1}^{2n}\frac{1}{k}$ How can I prove the following equation?:
$$s=\sum_{k=1}^{n}\frac{1}{4k^{2}-2k}=\sum_{k=n+1}^{2n}\frac{1}{k}$$
Simplifying both terms of the equation:
$$\sum_{k=1}^{n}\frac{1}{4k^{2}-2k} = \sum_{k=1}^{n}\frac{1}{(2k-1)2k} =(\sum_{k=1}^{n}\frac{1}{2k-1}-\sum_{k=1}^{n}\frac{1}{2k})$$
$$\sum_{k=n+1}^{2n}\frac{1}{k}=(\sum_{k=1}^{2n}\frac{1}{k}-\sum_{k=1}^{n}\frac{1}{k})=(\sum_{k=1}^{n}\frac{1}{2k-1}+\sum_{k=1}^{n}\frac{1}{2k}-\sum_{k=1}^{n}\frac{1}{k})$$
Now we have:
$$s=(\sum_{k=1}^{n}\frac{1}{2k-1}-\sum_{k=1}^{n}\frac{1}{2k})=(\sum_{k=1}^{n}\frac{1}{2k-1}\color{blue}{+}\sum_{k=1}^{n}\frac{1}{2k}\color{blue}{-\sum_{k=1}^{n}\frac{1}{k}})$$
How can I continue?
 [/INST] Let's do a cute induction on $n$: 
For $n = 1$, we have $\frac{1}{4 - 2} = \frac{1}{2}$, so the claim holds. 
Assume $n > 1$ and the claim holds for $n - 1$. Then $$\sum_{k = 1}^n \frac{1}{4k^2 - 2k} = \frac{1}{4n^2 - 2n} + \sum_{k = 1}^{n - 1} \frac{1}{4k^2 - 2k} = \frac{1}{4n^2 - 2n} + \sum_{k = n}^{2(n  -1)} \frac{1}{k}$$
by the induction hypothesis. So we only need to show that $\frac{1}{2n} + \frac{1}{2n - 1} - \frac{1}{n} = \frac{1}{4n^2 - 2n}$, which is just a direct computation. 
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