1. From the binomial expansion of \((1+x)^{10}\), the coefficient \(a_r\) is given by:
\[
a_r = \binom{10}{r}.
\]
2. The ratio of consecutive coefficients \(\frac{a_r}{a_{r-1}}\) is:
\[
\frac{a_r}{a_{r-1}} = \frac{\binom{10}{r}}{\binom{10}{r-1}} = \frac{10-r+1}{r}.
\]
3. Substituting this into \(\sum_{r=1}^{10} r^3 \left( \frac{a_r}{a_{r-1}} \right)^2\), we get:
\[
\sum_{r=1}^{10} r^3 \left( \frac{10-r+1}{r} \right)^2.
\]
4. Simplify the term inside the summation:
\[
\left( \frac{10-r+1}{r} \right)^2 = \frac{(10-r+1)^2}{r^2}.
\]
5. The summation becomes:
\[
\sum_{r=1}^{10} r^3 \cdot \frac{(10-r+1)^2}{r^2} = \sum_{r=1}^{10} r \cdot (10-r+1)^2.
\]
6. Expand \((10-r+1)^2\):
\[
(10-r+1)^2 = (11-r)^2 = 121 - 22r + r^2.
\]
7. Substituting this back, the summation becomes:
\[
\sum_{r=1}^{10} r \cdot (121 - 22r + r^2).
\]
8. Split the summation into three parts:
\[
\sum_{r=1}^{10} r \cdot 121 - \sum_{r=1}^{10} r \cdot 22r + \sum_{r=1}^{10} r \cdot r^2.
\]
9. Calculate each part:
- First part: \(\sum_{r=1}^{10} 121r = 121 \sum_{r=1}^{10} r = 121 \cdot \frac{10(10+1)}{2} = 121 \cdot 55 = 6655.\)
- Second part: \(\sum_{r=1}^{10} 22r^2 = 22 \sum_{r=1}^{10} r^2 = 22 \cdot \frac{10(10+1)(2 \cdot 10+1)}{6} = 22 \cdot 385 = 8470.\)
- Third part: \(\sum_{r=1}^{10} r^3 = \left( \frac{10(10+1)}{2} \right)^2 = 55^2 = 3025.\)
10. Combine the results:
\[
\sum_{r=1}^{10} r^3 \left( \frac{a_r}{a_{r-1}} \right)^2 = 6655 - 8470 + 3025 = 1210.
\]
Final Answer: 1210.
The key to solving this problem lies in simplifying the ratio of binomial coefficients and expanding the summation. Recognizing standard summation formulas for \(r\), \(r^2\), and \(r^3\) is crucial.