$$\frac{2^3 - 1^3}{1 \times 7} + \frac{4^3 - 3^3 + 2^3 - 1^3}{2 \times 11} + \frac{6^3 - 5^3 + 4^3 - 3^3 + 2^3 - 1^3}{3 \times 15} + \ldots + \frac{30^3 - 29^3 + 28^3 - 27^3 + \ldots + 2^3 - 1^3}{15 \times 63}$$ is equal to ______.

We need to evaluate the sum:

$$S = \sum_{r=1}^{15} \frac{(2r)^3 - (2r-1)^3 + (2r-2)^3 - (2r-3)^3 + \ldots + 2^3 - 1^3}{r(4r+3)}$$

First, we simplify the numerator of the r-th term, which is an alternating sum that can be paired as follows:

$$N_r = (2r)^3 - (2r-1)^3 + (2r-2)^3 - (2r-3)^3 + \ldots + 2^3 - 1^3 = \sum_{k=1}^{r} \left[(2k)^3 - (2k-1)^3\right].$$

To verify this pattern, observe that for r = 1 we have 2^3 - 1^3 = 7 with denominator 1 × 7 = 7, for r = 2 the numerator is 4^3 - 3^3 + 2^3 - 1^3 = 44 with denominator 2 × 11 = 22, and for r = 3 it becomes 6^3 - 5^3 + 4^3 - 3^3 + 2^3 - 1^3 = 135 with denominator 3 × 15 = 45.

We note that the denominators follow the pattern 1 × 7, 2 × 11, 3 × 15, …, r × (4r + 3), and in particular for r = 15 the denominator is 15 × 63 since 4(15) + 3 = 63.

Next, from the identity a^3 - b^3 = (a-b)(a^2 + ab + b^2) with a = 2k and b = 2k-1, we obtain

$$ (2k)^3 - (2k-1)^3 = 1 \cdot (4k^2 + 2k(2k-1) + (2k-1)^2) = 4k^2 + 4k^2 - 2k + 4k^2 - 4k + 1 = 12k^2 - 6k + 1. $$

Therefore, substituting into the sum and using standard formulas gives

$$N_r = \sum_{k=1}^{r} (12k^2 - 6k + 1) = 12 \cdot \frac{r(r+1)(2r+1)}{6} - 6 \cdot \frac{r(r+1)}{2} + r$$

$$= 2r(r+1)(2r+1) - 3r(r+1) + r = r\bigl[2(r+1)(2r+1) - 3(r+1) + 1\bigr]$$

$$= r\bigl[2(2r^2 + 3r + 1) - 3r - 3 + 1\bigr] = r\bigl[4r^2 + 6r + 2 - 3r - 2\bigr] = r\bigl[4r^2 + 3r\bigr] = r^2(4r + 3).$$

Substituting this result into the general term of the sum yields

$$ \frac{N_r}{r(4r+3)} = \frac{r^2(4r+3)}{r(4r+3)} = r. $$

Finally, summing over r from 1 to 15 gives

$$ S = \sum_{r=1}^{15} r = \frac{15 \times 16}{2} = 120. $$

Therefore, the answer is $$\boxed{120}$$.