The value of $\sum _{r=1}^{15} r^2\left(\frac{{ }^{15}{C}_r}{{ }^{15}{C}_{r-1}}\right)$ is equal to

We know that $\frac{{ }^n{C}_r}{{ }^n{C}_{r-1}}=\frac{n-r+1}{r}$

$\begin{array}{l}\sum _{r=1}^{15} r^2\left(\frac{{ }^{15}{C}_r}{{ }^{15}{C}_{r-1}}\right)=\sum _{r=1}^{15} r^2\left(\frac{15-r+1}{r}\right)=\sum _{r=1}^{15} r(16-r)\\ =16\sum _{r=1}^{15} r-\sum _{r=1}^{15} r^2\\ =\frac{16\times 15\times 16}{2}-\frac{15\times 16\times 31}{6}=680\end{array}$