Find the sum of $\sum_{r=1}^{n}r(^nC_r)^2$.
I am attempting it as follows $(1+x)^n=C_0+xC_1+x^2C_2+x^3C_3+⋯+x^nC_n$. On differentiating we get following relation $n(1+x)^{n−1}=\sum_{r=1}^n rx^{r−1}C_r$. Then multiply both sides by $x$. We get $nx(1+x)^{n−1}=\sum_{r=1}^n rx^rC_r$. Now I am considering multiplication of $(1+x)^n$ on both sides, and finding the sum of coefficients of $x^2,x^4,x^6,…,x^{2n}$ on the RHS.
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{r = 1}^{n}r\pars{^{n}C_{r}}^{2} & = \sum_{r = 0}^{n}r{n \choose r}{n \choose n - r} = \sum_{r = 0}^{n}r{n \choose r}\bracks{z^{n - r}}\pars{1 + z}^{n} \\[5mm] & = \bracks{z^{n}}\pars{1 + z}^{n} \sum_{r = 0}^{n}{n \choose r}r\, z^{r} = \bracks{z^{n}}\pars{1 + z}^{n}\ z\,\partiald{}{z} \sum_{r = 0}^{n}{n \choose r}z^{r} \\[5mm] & = \bracks{z^{n - 1}}\pars{1 + z}^{n}\,\partiald{\pars{1 + z}^{n}}{z} = n\bracks{z^{n - 1}}\pars{1 + z}^{2n - 1} \\[5mm] & = \bbx{n{2n - 1 \choose n - 1}} \end{align}