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?
 A: $$\sum_{n=1}^{10}\frac{1}{4n^2-1}=\frac12\sum_{n=1}^{10}\frac{1}{2n-1}-\frac{1}{2n+1}=\frac12\left(1-\frac{1}{21}\right)=\frac{10}{21}$$
A: Lemma to find the sum of $n$ terms of a series each term of which
is composed of the reciprocal of the product of $r$ factors in arithmetical progression,
the first factors of the several terms being in
the same arithmetical progression:
Write down the $nth$ term,
strike off a factor from the beginning,
divide by the number of factors so diminished and by the common difference,
change the sign and add a constant.
In our case, the general term is $T_n$=$$\frac 1 {(2n)^2-1}$$
Which is simply, $$\frac 1 {(2n-1)(2n+1)}$$
Applying the lemma, we get:
$$S_n=C-(\frac 1 2)(\frac 1 {2n+1})$$
$$S_n=C-\frac 1 {4n+2}$$
Where $C$ is a constant quantity.
Putting $n=1$ in general term, we get:
$$S_1=\frac 1 3=C- \frac 1 {4+2}$$
That gives $$C=\frac 1 2$$
Thus, $$S_n=\frac 1 2-\frac 1 {4n+2}$$
Putting $n=10$ gives:
$$S_{10}=\frac 1 2-\frac 1 {40+2}$$
$$S_{10}=\frac 1 2-\frac 1 {42}$$
$$Or, S_{10}=\frac {20} {42}=\frac {10} {21}$$
 Which is the desired answer.
The method is a bit, long... But it is great for the so-called "bashing" through such problems. If you wish to see a proof, I refer you to Hall and Knight's Higher Algebra.
(P.S. I did this long answer with this specific method just so I could get a hang of the typing system in StackExchange. So, please forgive me even if the method seems useless... I'm new here...)  
A: Alternatively to the telescoping sum decomposition, there is an easy pattern
$$\frac13$$
$$\frac13+\frac1{15}=\frac25$$
$$\frac13+\frac1{15}+\frac1{35}=\frac37$$
$$\frac13+\frac1{15}+\frac1{35}+\frac1{63}=\frac49$$
$$\cdots$$
