If a_r  is the coefficient of ${\mathrm{x}}^{10-\mathrm{r}}$  in the Binomial expansion of ${(1+\mathrm{x})}^{10}$ , then $\sum _{\mathrm{r}=1}^{10}{\mathrm{r}}^3{\left(\frac{{\mathrm{a}}_{\mathrm{r}}}{{\mathrm{a}}_{\mathrm{r}-1}}\right)}^2$  is equal to 

$\begin{array}{l}{\mathrm{a}}_{\mathrm{r}}= \mathrm{coefficient} \mathrm{of} {\mathrm{x}}^{10-\mathrm{r}} \mathrm{in} (1+\mathrm{x}{)}^{10}=\left({ }^{10}{\mathrm{C}}_{10-\mathrm{r}}\right){=}^{10}{\mathrm{C}}_{\mathrm{r}}\\ \sum _{\mathrm{r}=1}^{10} {\mathrm{r}}^3\cdot {\left(\frac{{\mathrm{a}}_{\mathrm{r}}}{{\mathrm{a}}_{\mathrm{r}-1}}\right)}^2=\sum {\mathrm{r}}^3{\left[\frac{{ }^{10}{\mathrm{C}}_{\mathrm{r}}}{{ }^{10}{\mathrm{C}}_{\mathrm{r}-1}}\right]}^2=\sum {\mathrm{r}}^3{\left(\frac{11-\mathrm{r}}{\mathrm{r}}\right)}^2\end{array}$
$=\sum \mathrm{r}(121-22\mathrm{r}+{\mathrm{r}}^2)$
$=121\times \frac{(10\times 11)}{2}-22\times \left(\frac{10\times 11\times 21}{6}\right)+{\left(\frac{10\times 11}{2}\right)}^2$
$=55\times [121-154+55]=55\times 22=120$