I'll shed some light on another way to do this using a slightly different method with partial fractions. Note that the way Marvis did it is generally a more effective method for more difficult expressions (say with several variables where systems of equations are nice to work with). $\frac{1}{{4n^2 - 1}} = \frac{1}{(2n+1)(2n-1)} = \frac{A}{2n+1} + \frac{B}{2n-1}$
We can cross multiply to get
$1 = A(2n-1) + B(2n+1)$
To easily find the values of $A$ and $B$, it would be nice if we were able to get it into a single variable linear equation rather than having to deal with systems of equations. This is the same as saying that we want to try to find a $n$ such that $A(2n-1) = 0$ and $B(2n+1) = 0$.
$2n-1 = 0 \implies 2n = 1 \implies n = \frac{1}{2}$ $2n + 1 = 0 \implies 2n = - 1 \implies n = -\frac{1}{2}$
So, let's try both of those values now.
Plugging in $n = \frac{1}{2}$ yields
$1 = 2B \iff B = \frac{1}{2}$
Plugging in $n = -\frac{1}{2}$ yields
$1 = -2A \iff A = -\frac{1}{2}$
Hence, $\frac{1}{{4n^2 - 1}} = \frac{1}{{(2n + 1)(2n - 1)}} = \frac{1}{{2(2n - 1)}} - \frac{1}{{2(2n + 1)}}$