If a _ r is the coefficient of x ^ 10 - r in the Binomial expansion...

Since, ${a}_{r}$ is the coefficient of ${x}^{10 - r}$ in the binomial expansion of ${\left(1 + x\right)}^{10}$, therefore

${a}_{r} = C10 - r 10 = Cr 10$

${a}_{r - 1} = Cr - 1 10$

So,

$\dfrac{{a}_{r}}{{a}_{r - 1}} = \dfrac{Cr 10}{Cr - 1 10}$

$⇒ \dfrac{{a}_{r}}{{a}_{r - 1}} = \dfrac{10 !}{r ! \times \left(10 - r\right) !} \times \dfrac{\left(r - 1\right) ! \times \left(11 - r\right) !}{10 !}$

$⇒ \dfrac{{a}_{r}}{{a}_{r - 1}} = \dfrac{11 - r}{r}$

Now,

$\mathop{∑}_{r = 1}^{10}{r}^{3}{\left(\dfrac{{a}_{r}}{{a}_{r - 1}}\right)}^{2} = \mathop{∑}_{r = 1}^{10}{r}^{3}{\left(\dfrac{11 - r}{r}\right)}^{2}$

$⇒ \mathop{∑}_{r = 1}^{10}{r}^{3}{\left(\dfrac{{a}_{r}}{{a}_{r - 1}}\right)}^{2} = \mathop{∑}_{r = 1}^{10}r{\left(11 - r\right)}^{2}$

$⇒ \mathop{∑}_{r = 1}^{10}{r}^{3}{\left(\dfrac{{a}_{r}}{{a}_{r - 1}}\right)}^{2} = \mathop{∑}_{r = 1}^{10}r\left(121 + {r}^{2} - 22r\right)$

$⇒ \mathop{∑}_{r = 1}^{10}{r}^{3}{\left(\dfrac{{a}_{r}}{{a}_{r - 1}}\right)}^{2} = \mathop{∑}_{r = 1}^{10}\left(121r + {r}^{3} - 22{r}^{2}\right)$

$⇒ \mathop{∑}_{r = 1}^{10}{r}^{3}{\left(\dfrac{{a}_{r}}{{a}_{r - 1}}\right)}^{2} = \dfrac{121 \times 10 \times 11}{2} + {\left(\dfrac{10 \times 11}{2}\right)}^{2} - \dfrac{22 \times 10 \times 11 \times 21}{6}$

$⇒ \mathop{∑}_{r = 1}^{10}{r}^{3}{\left(\dfrac{{a}_{r}}{{a}_{r - 1}}\right)}^{2} = 6655 + 3025 - 8470$

$⇒ \mathop{∑}_{r = 1}^{10}{r}^{3}{\left(\dfrac{{a}_{r}}{{a}_{r - 1}}\right)}^{2} = 1210$