Let $a_n=\sum _{r=0}^n \frac{1}{{\left({ }^n{C}_r\right)}^2}$ and $b_n=\sum _{r=0}^n \frac{(n-r)}{{\left({ }^n{C}_r\right)}^2}$

then $b_n$ equals 

We have

$b_n=\sum _{r=0}^n \frac{n-r}{{\left({ }^n{C}_r\right)}^2}=\sum _{r=0}^n \frac{n-r}{{\left({ }^n{C}_{n-r}\right)}^2}=\sum _{r=0}^n \frac{r}{{\left({ }^n{C}_r\right)}^2}$

Thus, 

$b_n+b_n=\sum _{r=0}^n \frac{n-r}{{\left(n_{C_r}\right)}^2}+\sum _{r=0}^n \frac{r}{{\left(n_{C_r}\right)}^2}=n\sum _{r=0}^n \frac{1}{{\left({ }^n{C}_r\right)}^2}=n{a}_n$

$\Rightarrow b_n=\frac{n}{2}{a}_n$