Question

Let $X={(}^{10}{C}_1{)}^2+{(}^{10}{C}_2{)}^2+3{(}^{10}{C}_3{)}^2+....+10{(}^{10}{C}_{10}{)}^2$, where ${}^{10}{C}_r,rϵ\{1,2,...,10\}$ denote binomial coefficients. Then, the value of $\frac{1}{1430}X$ is

Answer 1

$X=\sum _{r=0}^{n}r.{(}^n{C}_r{)}^2;n=10$
$\Rightarrow X=n.\sum _{r=0}^n{}^n{C}_r{.}^{n-1}{C}_{r-1}$

$\Rightarrow X=n.\sum _{r=0}^n{}^n{C}_{n-r}{.}^{n-1}{C}_{r-1}$

$\Rightarrow X=n{.}^{2n-1}{C}_{n-1};n=10$

$\Rightarrow X=10{.}^{19}{C}_9$

$\Rightarrow \frac{X}{1430}=\frac{1}{143}{.}^{19}{C}_9$

$=646$.