$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\sum_{k = 1}^{n}{1 \over n^{2} - 4\pars{k - 1}^{2}} ={\pi n\cot\pars{\pi n/2} - n\Psi\pars{n/2} + n\Psi\pars{3 n/2} + 2 \over 4 n^{2}}:\ {\large ?}}$
\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 $\pars{1}$.