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$\ds{\sum_{k = 1}^{n}{1 \over n^{2} - 4\pars{k - 1}^{2}}
={\pi n\cot\pars{\pi n/2} - n\Psi\pars{n/2} + n\Psi\pars{3 n/2} + 2
\over 4 n^{2}}:\ {\large ?}}$
\begin{align}
&\color{#c00000}{\sum_{k = 1}^{n}{1 \over n^{2} - 4\pars{k - 1}^{2}}}
=-\,{1 \over 4}\sum_{k = 0}^{n - 1}{1 \over k^{2} - n^{2}/4}
=-\,{1 \over 4}\sum_{k = 0}^{n - 1}{1 \over \pars{k + n/2}\pars{k - n/2}}
\\[3mm]&=-\,{1 \over 4}\sum_{k = 0}^{\infty}\bracks{%
{1 \over \pars{k + n/2}\pars{k - n/2}} - {1 \over \pars{k + 3n/2}\pars{k + n/2}}}
\\[3mm]&=-\,{1 \over 4}\,
{\Psi\pars{n/2} - \Psi\pars{-n/2} \over \pars{n/2} - \pars{-n/2}}
+ {1 \over 4}\,
{\Psi\pars{3n/2} - \Psi\pars{n/2} \over \pars{3n/2} - \pars{n/2}}
\end{align}
where $\ds{\Psi\pars{z}}$ is the
Digamma Function. See ${\bf\mbox{6.3.1}}$ in that link.
$$
\color{#c00000}{\sum_{k = 1}^{n}{1 \over n^{2} - 4\pars{k - 1}^{2}}}
={\Psi\pars{3n/2} - 2\Psi\pars{n/2} + \Psi\pars{-n/2} \over 4n}\tag{1}
$$
With formula ${\bf\mbox{6.3.7}}$ we'll get
$$
\Psi\pars{-\,{n \over 2}} = \Psi\pars{1 + {n \over 2}}
-\pi\cot\pars{\pi\bracks{-\,{n \over 2}}}
$$
Moreover, $\ds{\Psi\pars{1 + n/2} = \Psi\pars{n/2} + 1/\pars{n/2}}$ where we used
the identity ${\bf\mbox{6.3.5}}$
such that:
$$
\Psi\pars{-\,{n \over 2}}
=\Psi\pars{n \over 2} + {2 \over n} + \pi\cot\pars{\pi n \over 2}
$$
The final answer is found by replacing this result in expression $\pars{1}$.