<s>[INST] Find $ \frac{1}{2^2 –1} + \frac{1}{4^2 –1} + \frac{1}{6^2 –1} + \ldots + \frac{1}{20^2 –1} $ Find the following sum

$$
\frac{1}{2^2 –1} + \frac{1}{4^2 –1} + \frac{1}{6^2 –1} + \ldots + \frac{1}{20^2 –1}
$$

I am not able to find any short trick for it.
Is there any short trick or do we have to simplify and add it?
 [/INST] $$\sum_{n=1}^{10}\frac{1}{4n^2-1}=\frac12\sum_{n=1}^{10}\frac{1}{2n-1}-\frac{1}{2n+1}=\frac12\left(1-\frac{1}{21}\right)=\frac{10}{21}$$
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