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

The coefficient of $$x^{10-r}$$ in $$(1+x)^{10}$$ is $$a_r = \binom{10}{10-r}$$.

Similarly, $$a_{r-1} = \binom{10}{11-r}$$.

$$\frac{a_r}{a_{r-1}} = \frac{\binom{10}{10-r}}{\binom{10}{11-r}} = \frac{\frac{10!}{(10-r)!r!}}{\frac{10!}{(11-r)!(r-1)!}} = \frac{(11-r)!(r-1)!}{(10-r)!r!} = \frac{11-r}{r}$$

So $$\left(\frac{a_r}{a_{r-1}}\right)^2 = \frac{(11-r)^2}{r^2}$$

$$\sum_{r=1}^{10} r^3 \cdot \frac{(11-r)^2}{r^2} = \sum_{r=1}^{10} r(11-r)^2$$

$$= \sum_{r=1}^{10} r(121 - 22r + r^2) = 121\sum r - 22\sum r^2 + \sum r^3$$

Using standard formulas: $$\sum_{r=1}^{10} r = 55$$, $$\sum_{r=1}^{10} r^2 = 385$$, $$\sum_{r=1}^{10} r^3 = 3025$$.

$$= 121(55) - 22(385) + 3025 = 6655 - 8470 + 3025 = 1210$$

The correct answer is Option 2: 1210.