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?

Alternatively to the telescoping sum decomposition, there is an easy pattern
$$\frac13$$
$$\frac13+\frac1{15}=\frac25$$
$$\frac13+\frac1{15}+\frac1{35}=\frac37$$
$$\frac13+\frac1{15}+\frac1{35}+\frac1{63}=\frac49$$
$$\cdots$$
