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$\ds{\sum_{k = 0}^{n}{n \choose k}^{2} =
{n + 1 \over n}\sum_{k = 1}^{n}{n \choose k}{n \choose k - 1}:\ ?}$.
\begin{align}
&\color{#f00}{{n + 1 \over n}\sum_{k = 1}^{n}{n \choose k}{n \choose k - 1}} =
{n + 1 \over 2n}\sum_{k = 1}^{n}\braces{%
\bracks{{n \choose k} + {n \choose k - 1}}^{2} - {n \choose k}^{2} -
{n \choose k - 1}^{2}}
\\[5mm] = &\
{n + 1 \over 2n}\sum_{k = 1}^{n}{n + 1 \choose k}^{2} -
{n + 1 \over 2n}\sum_{k = 1}^{n}{n \choose k}^{2} -
{n + 1 \over 2n}\sum_{k = 1}^{n}{n \choose k - 1}^{2}
\quad\pars{\begin{array}{l}Pascal\ Triangle\ Identity\
\\ \mbox{in the first term}\end{array}}
\\[5mm] = &\
{n + 1 \over 2n}\bracks{-1 + \sum_{k = 0}^{n + 1}{n + 1 \choose k}^{2} - 1} -
{n + 1 \over 2n}\bracks{-1 + \sum_{k = 0}^{n}{n \choose k}^{2}} -
{n + 1 \over 2n}\bracks{\sum_{k = 0}^{n}{n \choose k}^{2} - 1}
\\[5mm] = &\
\color{#f00}{{n + 1 \over 2n}\sum_{k = 0}^{n + 1}{n + 1 \choose k}^{2} -
{n + 1 \over n}\sum_{k = 0}^{n}{n \choose k}^{2}}\label{1}\tag{1}
\end{align}
\begin{equation}
\mbox{We'll use the}\ well\ known\ identity\quad
\sum_{j = 0}^{m}{m \choose j}^{2} = {2m \choose m}\label{2}\tag{2}
\end{equation}
Expression \eqref{1} becomes:
\begin{align}
&\color{#f00}{{n + 1 \over n}\sum_{k = 1}^{n}{n \choose k}{n \choose k - 1}} =
{n + 1 \over 2n}{2n + 2 \choose n + 1} -
{n + 1 \over n}\sum_{k = 0}^{n}{n \choose k}^{2}
\\[5mm] = &\
{n + 1 \over 2n}\bracks{{\pars{2n + 2}\pars{2n + 1} \over
\pars{n + 1}\pars{n + 1}}{2n \choose n}} -
{n + 1 \over n}\sum_{k = 0}^{n}{n \choose k}^{2} =
{2n + 1 \over n}\ \overbrace{\sum_{k = 0}^{n}{n \choose k}^{2}}
^{\ds{\mbox{see}\ \eqref{2}}}\ -\
{n + 1 \over n}\sum_{k = 0}^{n}{n \choose k}^{2}
\\[5mm] = &\
\color{#f00}{\sum_{k = 0}^{n}{n \choose k}^{2}}
\end{align}