Question:

Step 1: Determine the coefficient $a_r$. The binomial expansion of $(1+x)^{10}$ is given by the general term $T_{k+1} = \binom{10}{k}x^k$. We are given that $a_r$ is the coefficient of $x^{10-r}$. By comparing the power of $x$, we set $k = 10-r$. Therefore, the coefficient $a_r$ is given by: $$a_r = \binom{10}{10-r}$$ Using the property $\binom{n}{k} = \binom{n}{n-k}$, we can simplify $a_r$: $$a_r = \binom{10}{10-(10-r)} = \binom{10}{r}$$ Step 2: Calculate the ratio $\dfrac{a_r}{a_{r-1}}$. Using the expression for $a_r$ from Step 1, we can find $a_{r-1}$: $$a_r = \binom{10}{r} \quad \text{and} \quad a_{r-1} = \binom{10}{r-1}$$ Now, we compute the ratio $\dfrac{a_r}{a_{r-1}}$: $$\dfrac{a_r}{a_{r-1}} = \dfrac{\binom{10}{r}}{\binom{10}{r-1}}$$ Expanding the binomial coefficients: $$\dfrac{\binom{10}{r}}{\binom{10}{r-1}} = \dfrac{\frac{10!}{r!(10-r)!}}{\frac{10!}{(r-1)!(10-(r-1))!}} = \dfrac{\frac{10!}{r!(10-r)!}}{\frac{10!}{(r-1)!(11-r)!}}$$ $$= \dfrac{10!}{r!(10-r)!} \times \dfrac{(r-1)!(11-r)!}{10!}$$ $$= \dfrac{(r-1)!(11-r)!}{r!(10-r)!}$$ We know that $r! = r \times (r-1)!$ and $(11-r)! = (11-r) \times (10-r)!$. Substituting these into the expression: $$= \dfrac{(r-1)! \times (11-r) \times (10-r)!}{r \times (r-1)! \times (10-r)!} = \dfrac{11-r}{r}$$ Step 3: Compute the term $r^3\left(\dfrac{a_r}{a_{r-1}}\right)^2$. Substitute the ratio found in Step 2 into the given expression: $$r^3\left(\dfrac{a_r}{a_{r-1}}\right)^2 = r^3\left(\dfrac{11-r}{r}\right)^2$$ $$= r^3 \cdot \dfrac{(11-r)^2}{r^2} = r(11-r)^2$$ Expand the term $(11-r)^2$: $$= r(121 - 22r + r^2) = 121r - 22r^2 + r^3$$ Step 4: Evaluate the sum $\displaystyle\sum_{r=1}^{10}r^3\left(\dfrac{a_r}{a_{r-1}}\right)^2$. Substitute the simplified term from Step 3 into the summation: $$\sum_{r=1}^{10}(121r - 22r^2 + r^3)$$ This can be split into three separate summations: $$= 121\sum_{r=1}^{10}r - 22\sum_{r=1}^{10}r^2 + \sum_{r=1}^{10}r^3$$ Now, we use the standard summation formulas for $n=10$: $\sum_{r=1}^{n}r = \dfrac{n(n+1)}{2}$ $\sum_{r=1}^{n}r^2 = \dfrac{n(n+1)(2n+1)}{6}$ $\sum_{r=1}^{n}r^3 = \left(\dfrac{n(n+1)}{2}\right)^2$ For $n=10$: $\sum_{r=1}^{10}r = \dfrac{10(10+1)}{2} = \dfrac{10 \times 11}{2} = 55$ $\sum_{r=1}^{10}r^2 = \dfrac{10(10+1)(2 \times 10+1)}{6} = \dfrac{10 \times 11 \times 21}{6} = \dfrac{2310}{6} = 385$ $\sum_{r=1}^{10}r^3 = \left(\dfrac{10(10+1)}{2}\right)^2 = (55)^2 = 3025$ Substitute these values back into the sum: $$= 121(55) - 22(385) + 3025$$ $$= 6655 - 8470 + 3025$$ $$= 9680 - 8470$$ $$= 1210$$ The final answer is $\boxed{1210}$.