Let  $X={\left({ }^{10}{C}_1\right)}^2+2{\left({ }^{10}{C}_2\right)}^2+3{\left({ }^{10}{C}_3\right)}^2+\dots$ 

$+10{\left({ }^{10}{C}_{10}\right)}^2,$ where ${ }^{10}{C}_r,r=\{1,2,\dots ,10\}$  denote binomial

coefficients. Then the value of   $\frac{X}{1430}$ is 

We have 

  $\begin{array}{l}X={\left({ }^{10}{C}_1\right)}^2+2{\left({ }^{10}{C}_2\right)}^2+3{\left({ }^{10}{C}_3\right)}^2+\dots +10{\left({ }^{10}{C}_{10}\right)}^2\\ \Rightarrow X=\sum _{r=1}^{10} r{\left({ }^{10}{C}_r\right)}^2\\ \Rightarrow X=\sum _{r=1}^{10} r^{10}{C}_r{\cdot }^{10}{C}_r=\sum _{r=1}^2 r\cdot {\frac{10}{r}}^9{C_{r-1}}^{10}{C}_{10-r}\\ \Rightarrow X=10\sum _{r=1}^{10}{ }^9{C}_{r-1}{\cdot }^{10}{C}_{10-r}\end{array}$

  $\Rightarrow X=10$  Coeffieient  of $x^9$ in  $(1+x{)}^9(1+x{)}^{10}$

  $\Rightarrow X=10$ Coeffieient  of $x^9$in  $(1+x{)}^{19}$ 

  $\begin{array}{l}\Rightarrow X=10{\times }^{19}{C}_9\\ \therefore \frac{X}{1430}=\frac{10{\times }^{19}{C}_9}{1430}=646\end{array}$