Key Concepts and Formulas

• Sum of the first $n$ natural numbers: $\sum_{k=1}^n k = \frac{n(n+1)}{2}$
• Sum of the cubes of the first $n$ natural numbers: $\sum_{k=1}^n k^3 = \left(\frac{n(n+1)}{2}\right)^2$
• Sum of the squares of the first $n$ natural numbers: $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$
• Properties of Summation: Linearity of summation, i.e., $\sum (a_n \pm b_n) = \sum a_n \pm \sum b_n$ and $\sum c \cdot a_n = c \sum a_n$.

Step-by-Step Solution

The problem asks us to evaluate the sum: $S = \left( {1 + {{{1^3} + {2^3}} \over {1 + 2}} + {{{1^3} + {2^3} + {3^3}} \over {1 + 2 + 3}} + ...... + {{{1^3} + {2^3} + {3^3} + ... + {{15}^3}} \over {1 + 2 + 3 + ... + 15}}} \right) - {1 \over 2}\left( {1 + 2 + 3 + ... + 15} \right)$

Step 1: Express the sum in summation notation and simplify the general term of the series. The given expression consists of two parts. The first part is a series where the $n$-th term, let's call it $T_n$, is given by the ratio of the sum of the first $n$ cubes to the sum of the first $n$ natural numbers. The series runs from $n=1$ to $n=15$. The second part is a simple arithmetic series multiplied by $-1/2$. We can write the sum as: $S = \sum_{n=1}^{15} T_n - \frac{1}{2} \sum_{n=1}^{15} n$ where $T_n = \frac{\sum_{k=1}^n k^3}{\sum_{k=1}^n k}$.

The crucial insight for simplifying this problem is to first simplify the general term $T_n$. Using the formulas for the sum of cubes and the sum of natural numbers: $T_n = \frac{\left(\frac{n(n+1)}{2}\right)^2}{\frac{n(n+1)}{2}}$ This simplifies to: $T_n = \frac{n(n+1)}{2}$ This simplification is vital as it converts a complex fractional term into a simple quadratic expression.

Step 2: Substitute the simplified general term back into the sum and perform further simplification. Now, substitute $T_n = \frac{n(n+1)}{2}$ back into the expression for $S$: $S = \sum_{n=1}^{15} \frac{n(n+1)}{2} - \frac{1}{2} \sum_{n=1}^{15} n$ Expand the term inside the first summation: $S = \sum_{n=1}^{15} \frac{n^2 + n}{2} - \frac{1}{2} \sum_{n=1}^{15} n$ Using the linearity property of summation, we can split the first summation and factor out the constant $1/2$: $S = \frac{1}{2} \left( \sum_{n=1}^{15} n^2 + \sum_{n=1}^{15} n \right) - \frac{1}{2} \sum_{n=1}^{15} n$ Now, distribute the $1/2$: $S = \frac{1}{2} \sum_{n=1}^{15} n^2 + \frac{1}{2} \sum_{n=1}^{15} n - \frac{1}{2} \sum_{n=1}^{15} n$ We can see that the terms $\frac{1}{2} \sum_{n=1}^{15} n$ cancel each other out. This leaves us with: $S = \frac{1}{2} \sum_{n=1}^{15} n^2$ This significant simplification shows that the original complex sum reduces to a simple sum of squares.

Step 3: Evaluate the remaining summation. We need to calculate $S = \frac{1}{2} \sum_{n=1}^{15} n^2$. We use the formula for the sum of the squares of the first $N$ natural numbers, $\sum_{n=1}^N n^2 = \frac{N(N+1)(2N+1)}{6}$, with $N=15$. $\sum_{n=1}^{15} n^2 = \frac{15(15+1)(2 \cdot 15 + 1)}{6}$ $\sum_{n=1}^{15} n^2 = \frac{15(16)(31)}{6}$ Now, we perform the calculation: $\sum_{n=1}^{15} n^2 = \frac{15 \cdot 16 \cdot 31}{6} = \frac{(3 \cdot 5) \cdot (2 \cdot 8) \cdot 31}{2 \cdot 3}$ Cancel out the common factors of 2 and 3: $\sum_{n=1}^{15} n^2 = 5 \cdot 8 \cdot 31$ $\sum_{n=1}^{15} n^2 = 40 \cdot 31$ $\sum_{n=1}^{15} n^2 = 1240$ Finally, substitute this value back into the expression for $S$: $S = \frac{1}{2} (1240)$ $S = 620$

Common Mistakes & Tips

• Direct Calculation: Attempting to calculate each term of the series individually before summing would be extremely time-consuming and error-prone. Always look for a general term and a simplification strategy.
• Formula Errors: Ensure accurate recall and application of the summation formulas for $k$, $k^2$, and $k^3$.
• Algebraic Mistakes: Careful manipulation of fractions and summation properties is crucial to avoid errors during simplification.

Summary

The problem was solved by first expressing the given sum in summation notation. The key step was simplifying the general term of the series, which involved using the formulas for the sum of cubes and the sum of natural numbers. This simplification transformed the complex general term into a simple quadratic expression. Subsequently, by applying the linearity of summation and observing cancellations, the entire sum was reduced to a single summation of squares. Finally, the formula for the sum of squares was used to calculate the result.

The final answer is $\boxed{620}$.