1. Key Concepts and Formulas

   • Difference of Cubes: $a^3 - b^3 = (a-b)(a^2+ab+b^2)$.
   • Summation of Powers: $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$, $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$, $\sum_{k=1}^{n} c = cn$.
   • Arithmetic Progression: $a_n = a_1 + (n-1)d$.
   • Pattern Recognition and Algebraic Simplification for series terms.

2. Step-by-Step Solution

Step 1: Identify the structure of the $n$-th term ($T_n$) of the series.

Let's analyze the given terms: The first term is $T_1 = \frac{2^3-1^3}{1 \times 7}$. The second term is $T_2 = \frac{4^3-3^3+2^3-1^3}{2 \times 11}$. The third term is $T_3 = \frac{6^3-5^3+4^3-3^3+2^3-1^3}{3 \times 15}$. The series continues until the $15$-th term, which has $15$ in the denominator. This suggests that the series has $15$ terms in total, indexed from $n=1$ to $n=15$.

• Numerator ($N_n$): The numerator of the $n$-th term is a sum of $n$ pairs of differences of cubes. Each pair is of the form $(2k)^3 - (2k-1)^3$. For $T_1$ ($n=1$): $(2 \cdot 1)^3 - (2 \cdot 1 - 1)^3 = 2^3 - 1^3$. For $T_2$ ($n=2$): $(4^3 - 3^3) + (2^3 - 1^3) = \sum_{k=1}^{2} [(2k)^3 - (2k-1)^3]$. For $T_3$ ($n=3$): $(6^3 - 5^3) + (4^3 - 3^3) + (2^3 - 1^3) = \sum_{k=1}^{3} [(2k)^3 - (2k-1)^3]$. Thus, the general numerator is $N_n = \sum_{k=1}^{n} \left[ (2k)^3 - (2k-1)^3 \right]$.

• Denominator ($D_n$): The denominator of the $n$-th term has two factors. The first factor is $n$. The second factor forms an arithmetic progression: $7, 11, 15, \ldots$. The first term is $a_1 = 7$ and the common difference is $d = 11-7 = 4$. The $n$-th term of this AP is $a_n = a_1 + (n-1)d = 7 + (n-1)4 = 7 + 4n - 4 = 4n+3$. We can verify this for the last term: for $n=15$, the second factor is $4(15)+3 = 60+3 = 63$, which matches the given last term. So, the general denominator is $D_n = n(4n+3)$.

Therefore, the $n$-th term of the series is $T_n = \frac{\sum_{k=1}^{n} \left[ (2k)^3 - (2k-1)^3 \right]}{n(4n+3)}$.

Step 2: Simplify the expression inside the summation in the numerator.

We use the difference of cubes formula $a^3 - b^3 = (a-b)(a^2+ab+b^2)$ with $a=2k$ and $b=2k-1$. $a-b = (2k) - (2k-1) = 1$. So, $(2k)^3 - (2k-1)^3 = (1) \left[ (2k)^2 + (2k)(2k-1) + (2k-1)^2 \right]$. Expanding the terms: $(2k)^2 = 4k^2$. $(2k)(2k-1) = 4k^2 - 2k$. $(2k-1)^2 = (2k)^2 - 2(2k)(1) + 1^2 = 4k^2 - 4k + 1$. Summing these: $(2k)^3 - (2k-1)^3 = 4k^2 + (4k^2 - 2k) + (4k^2 - 4k + 1) = 12k^2 - 6k + 1$.

Step 3: Calculate the general numerator ($N_n$) by summing the simplified expression.

Now, we sum the simplified expression from $k=1$ to $n$: $N_n = \sum_{k=1}^{n} (12k^2 - 6k + 1)$. Using the linearity of summation and the standard sum formulas: $N_n = 12 \sum_{k=1}^{n} k^2 - 6 \sum_{k=1}^{n} k + \sum_{k=1}^{n} 1$. Substitute the formulas: $N_n = 12 \left( \frac{n(n+1)(2n+1)}{6} \right) - 6 \left( \frac{n(n+1)}{2} \right) + n$. Simplify: $N_n = 2n(n+1)(2n+1) - 3n(n+1) + n$. Factor out $n$: $N_n = n [2(n+1)(2n+1) - 3(n+1) + 1]$. Expand and simplify inside the brackets: $N_n = n [2(2n^2 + 3n + 1) - 3n - 3 + 1]$. $N_n = n [4n^2 + 6n + 2 - 3n - 2]$. $N_n = n [4n^2 + 3n]$. Factor out $n$ again: $N_n = n \cdot n(4n+3) = n^2(4n+3)$.

Step 4: Simplify the general term ($T_n$).

Substitute the simplified numerator $N_n$ back into the expression for $T_n$: $T_n = \frac{n^2(4n+3)}{n(4n+3)}$. Since $n$ ranges from $1$ to $15$, $n \neq 0$ and $4n+3 \neq 0$. We can cancel the common factors $n$ and $(4n+3)$: $T_n = n$. This is a surprisingly simple form for the $n$-th term.

Step 5: Calculate the total sum of the series.

The series has $15$ terms, from $n=1$ to $n=15$. The sum $S$ is given by: $S = \sum_{n=1}^{15} T_n$. Since $T_n = n$: $S = \sum_{n=1}^{15} n$. Using the formula for the sum of the first $N$ natural numbers, $\sum_{n=1}^{N} n = \frac{N(N+1)}{2}$, with $N=15$: $S = \frac{15(15+1)}{2}$. $S = \frac{15 \times 16}{2}$. $S = 15 \times 8$. $S = 120$.

1. Common Mistakes & Tips

   • Incorrectly identifying the pattern: Double-check the structure of the numerator and denominator for the general term with the first few given terms.
   • Algebraic errors in simplification: Be meticulous when expanding and combining terms, especially with the difference of cubes and summation formulas.
   • Misinterpreting the number of terms: The last term's denominator usually provides a clue to the total number of terms in the series.

2. Summary The problem involves a series where each term's numerator is a sum of differences of cubes, and the denominator follows a pattern related to the term number. By carefully identifying the general $n$-th term, simplifying the numerator using the difference of cubes identity and summation formulas, and then cancelling common factors with the denominator, we found that the $n$-th term simplifies to $n$. The sum of the series is then the sum of the first $15$ natural numbers, which is $120$.

The final answer is $\boxed{120}$.