Somewhat challenging binomial identity

We write le LHS term as $$ \eqalign{ & L(n) = \sum\limits_{k = 1}^n {{n \over 4}{{\left( { - 16} \right)^{\,k} } \over {k\left( {n + k} \right)}} \left( \matrix{ n + k \cr 2k \cr} \right)\left( \matrix{ 2k \cr k \cr} \right)^{\, - \,1} } = \cr & = \sum\limits_{1\, \le \,k} {{n \over 4}{{\left( { - 16} \right)^{\,k} } \over {k\left( {n + k} \right)}} \left( \matrix{ n + k \cr n - k \cr} \right)\left( \matrix{ 2k \cr k \cr} \right)^{\, - \,1} } = \cr & = \sum\limits_{0\, \le \,k} { - {{4n\left( { - 16} \right)^{\,k} } \over {\left( {k + 1} \right)\left( {n + k + 1} \right)}} \left( \matrix{ n + k + 1 \cr n - k - 1 \cr} \right)\left( \matrix{ 2k + 2 \cr k + 1 \cr} \right)^{\, - \,1} } = \cr & = \sum\limits_{0\, \le \,k} {T(k,n)} \cr} $$ so to get rid of the upper bound in the sum.

Then we reshape $T(k,n)$ so to render it more manageable by making use of the gamma duplication formula $$ \eqalign{ & T(k,n) = \cr & = {{ - \,4n\left( { - 16} \right)^{\,k} } \over {\left( {k + 1} \right)\left( {n + k + 1} \right)}} \left( \matrix{ n + k + 1 \cr n - k - 1 \cr} \right)\left( \matrix{ 2k + 2 \cr k + 1 \cr} \right)^{\, - \,1} \quad \left| {\,0 \le k \le n - 1} \right.\quad = \cr & = - \,4n{{\Gamma \left( {n + 2 + k} \right)} \over {\left( {n + 1 + k} \right)\Gamma \left( {n - k} \right)\Gamma \left( {2k + 3} \right)}} {{\Gamma \left( {k + 2} \right)^{\,2} } \over {\left( {k + 1} \right)\Gamma \left( {2k + 3} \right)}}\left( { - 16} \right)^{\,k} = \cr & = - \,4n{{\Gamma \left( {n + 1 + k} \right)} \over {\Gamma \left( {n - k} \right)}} {{\Gamma \left( {k + 2} \right)\Gamma \left( {k + 1} \right)} \over {\Gamma \left( {2k + 3} \right)^{\,2} }}\left( { - 16} \right)^{\,k} = \cr & = - \,4n{{\Gamma \left( {n + 1 + k} \right)} \over {\Gamma \left( {n - k} \right)}} {{\Gamma \left( {3/2} \right)^{\,2} \Gamma \left( {k + 2} \right)\Gamma \left( {k + 1} \right)} \over {\Gamma \left( {k + 3/2} \right)^{\,2} \Gamma \left( {k + 2} \right)^{\,2} }}{{\left( { - 16} \right)^{\,k} } \over {4^{\,2\,k + 1} }} = \cr & = - \,n{{\Gamma \left( {n + 1 + k} \right)} \over {\Gamma \left( {n - k} \right)}} {{\Gamma \left( {3/2} \right)^{\,2} \Gamma \left( {k + 1} \right)} \over {\Gamma \left( {k + 3/2} \right)^{\,2} \Gamma \left( {k + 2} \right)}} \left( { - 1} \right)^{\,k} = \cr & = - n^{\,2} {{{{\Gamma \left( {n + 1 + k} \right)} \over {\Gamma \left( {n + 1} \right)}}} \over {{{\Gamma \left( {n - k} \right)} \over {\Gamma \left( n \right)}}}}{1 \over {{{\Gamma \left( {k + 3/2} \right)} \over {\Gamma \left( {3/2} \right)^{\,2} }}^{\,2} }}{{\left( { - 1} \right)^{\,k} } \over {\left( {k + 1} \right)}} = \cr & = - n^{\,2} {{\left( {n + 1} \right)^{\,\overline {\,k\,} } } \over {n^{\,\overline {\, - \,k\,} } }} {1 \over {\left( {3/2} \right)^{\,\overline {\,k\,} } \left( {3/2} \right)^{\,\overline {\,k\,} } }} {{\left( { - 1} \right)^{\,k} } \over {\left( {k + 1} \right)}} = \cr & = - n^{\,2} {{\left( {n - 1} \right)^{\,\underline {\,k\,} } \left( {n + 1} \right)^{\,\overline {\,k\,} } } \over {\left( {3/2} \right)^{\,\overline {\,k\,} } \left( {3/2} \right)^{\,\overline {\,k\,} } }}{{\left( { - 1} \right)^{\,k} } \over {\left( {k + 1} \right)}} = \cr & = - n^{\,2} {{\left( { - n + 1} \right)^{\,\overline {\,k\,} } \left( {n + 1} \right)^{\,\overline {\,k\,} } } \over {\left( {3/2} \right)^{\,\overline {\,k\,} } \left( {3/2} \right)^{\,\overline {\,k\,} } }}{1 \over {\left( {k + 1} \right)}} = \cr & = {{\left( { - n} \right)^{\,\overline {\,k + 1\,} } n^{\,\overline {\,k + 1\,} } } \over {\left( {3/2} \right)^{\,\overline {\,k\,} } \left( {3/2} \right)^{\,\overline {\,k\,} } }}{1 \over {\left( {k + 1} \right)}} \cr} $$

where the single steps should result quite clear.

We do not attempt to go through the Hypergeometric at this point, which looks complicate.
Instead we go and take the Forward Difference in $n$ $$ \Delta _{\,n} L(n) = \sum\limits_{0\, \le \,k} {\Delta _{\,n} T(k,n)} $$ taking advantage of not having the upper bound.

Now $$ \eqalign{ & \Delta _{\,n} \left( {\left( { - n} \right)^{\,\overline {\,k + 1\,} } n^{\,\overline {\,k + 1\,} } } \right) = \cr & = \left( { - n - 1} \right)^{\,\overline {\,k + 1\,} } \left( {n + 1} \right)^{\,\overline {\,k + 1\,} } - \left( { - n} \right)^{\,\overline {\,k + 1\,} } n^{\,\overline {\,k + 1\,} } = \cr & = \left( { - n - 1} \right)\left( { - n} \right)^{\,\overline {\,k\,} } \left( {n + 1} \right)^{\,\overline {\,k\,} } \left( {n + k + 1} \right) - \left( { - n} \right)^{\,\overline {\,k\,} } \left( { - n + k} \right)n\left( {n + 1} \right)^{\,\overline {\,k\,} } = \cr & = - \left( {2n + 1} \right)\left( {k + 1} \right)\left( { - n} \right)^{\,\overline {\,k\,} } \left( {n + 1} \right)^{\,\overline {\,k\,} } \cr} $$ which provides a $(k+1)$ factor, wishfully expected from the Rising Factorials to cancel the disturbing one in the previous derivation

Therefore $$ \eqalign{ & \Delta _{\,n} L(n) = - \left( {2n + 1} \right)\sum\limits_{0\, \le \,k} {{{\left( { - n} \right)^{\,\overline {\,k\,} } \left( {n + 1} \right)^{\,\overline {\,k\,} } 1^{\,\overline {\,k\,} } } \over {\left( {3/2} \right)^{\,\overline {\,k\,} } \left( {3/2} \right)^{\,\overline {\,k\,} } }}{1 \over {k!}}} = \cr & = - \left( {2n + 1} \right){}_3F_{\,2} \left( {\left. {\matrix{ { - n,\;1,\;n + 1} \cr {3/2,3/2} \cr } \;} \right|\;1} \right) \cr} $$ and we are lucky enough that the factors allow to apply the Saalschütz's theorem $$ \eqalign{ & \Delta _{\,n} L(n) = - \left( {2n + 1} \right){{\left( {3/2 - 1} \right)^{\,\overline {\,n\,} } \left( {3/2 - n - 1} \right)^{\,\overline {\,n\,} } } \over {\left( {3/2} \right)^{\,\overline {\,n\,} } \left( {3/2 - 1 - n - 1} \right)^{\,\overline {\,n\,} } }} = \cr & = - \left( {2n + 1} \right){{\left( {1/2} \right)\left( { - 1/2} \right)} \over {\left( {1/2 + n} \right)\left( { - n - 1/2} \right)}} = \cr & = {1 \over {\left( { - 2n - 1} \right)}} \cr} $$

In conclusion, indicating with $R(n)$ the RHS of the identity to demonstrate, we have $$ \left\{ \matrix{ L(0) = R(0) = 0 \hfill \cr \Delta _{\,n} L(n) = \Delta _{\,n} R(n) = - {1 \over {\left( {2n + 1} \right)}} \hfill \cr} \right. $$

and the thesis is proved

Note: There is an interesting integral representation of (1), namely

\begin{align*} \frac{n}{4}\sum_{k=1}^n\frac{(-16)^k}{k(n+k)}\binom{n+k}{2k}\binom{2k}{k}^{-1} \color{blue}{=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\frac{\cos(2nz)-1}{\sin z}\,dz =}\sum_{k=1}^n\frac{1}{1-2k} \end{align*}

The validity of this equality-chain is shown in this MSE post.