How do I find the following finite sum? 
Let $n \in \Bbb N$ be fixed. Let $C_r = \binom n r\ $ for $0 \leq r \leq n.$ Evaluate $$C_0^2 + 3 C_1^2 + \cdots + (2n+1) C_n^2.$$

Any hint in this regard will be highly appreciated. Thank you very much.
 A: Using $k\binom{n}{k}=n\binom{n-1}{k-1}$ for $k>0$, and Chu-Vandermonde identity,
\begin{align}\sum_{k=0}^{n}(2k+1)\binom{n}{k}^2&=2n\sum_{k=1}^{n}\binom{n-1}{k-1}\binom{n}{k}+\sum_{k=0}^{n}\binom{n}{k}^2\\&=2n\sum_{k=0}^{n-1}\binom{n-1}{k}\binom{n}{n-1-k}+\sum_{k=0}^{n}\binom{n}{k}\binom{n}{n-k}\\&=2n\binom{2n-1}{n-1}+\binom{2n}{n}=\color{blue}{(n+1)\binom{2n}{n}}.\end{align}
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
&\bbox[5px,#ffd]{\sum_{r = 0}^{n}\pars{2r + 1}{n \choose r}^{2}} =
\sum_{r = 0}^{n}\pars{2r + 1}{n \choose r}{n \choose n - r}
\\[5mm] = &\
\sum_{r = 0}^{n}\pars{2r + 1}{n \choose r}\bracks{z^{n - r}}
\pars{1 + z}^{n} =
\bracks{z^{n}}
\pars{1 + z}^{n}\sum_{r = 0}^{n}z^{r}\pars{2r + 1}{n \choose r}
\\[5mm] = &\
\bracks{z^{n}}
\pars{1 + z}^{n}\pars{2z\,\partiald{}{z} + 1}\sum_{r = 0}^{n}{n \choose r}z^{r}
\\[5mm] = &\
\bracks{z^{n}}
\pars{1 + z}^{n}\pars{2z\,\partiald{}{z} + 1}\pars{1 + z}^{n}
\\[5mm] = &\
\bracks{z^{n}}
\pars{1 + z}^{n}\bracks{2nz\pars{1 + z}^{n - 1} + \pars{1 + z}^{n}}
\\[5mm] = &\
2n\bracks{z^{n - 1}}\pars{1 + z}^{2n - 1} +
\bracks{z^{n}}\pars{1 + z}^{2n} =
2n{2n - 1 \choose n - 1} + {2n \choose n}
\\[5mm] = &\
n\
\underbrace{\bracks{{2n \over n}\,{\pars{2n - 1}! \over
\pars{n - 1}!\, n!}}}_{\ds{2n \choose n}}\ +\ {2n \choose n}
\\[5mm] = &\ \bbx{\pars{n + 1}{2n \choose n}} \\ &
\end{align}
