Key Concepts and Formulas

• Sum of the first $r$ natural numbers: $\sum_{k=1}^r k = \frac{r(r+1)}{2}$
• Sum of the squares of the first $r$ natural numbers: $\sum_{k=1}^r k^2 = \frac{r(r+1)(2r+1)}{6}$
• Sum of the cubes of the first $r$ natural numbers: $\sum_{k=1}^r k^3 = \left(\frac{r(r+1)}{2}\right)^2$

Step-by-Step Solution

Step 1: Determine the General Term ($T_r$) of the Series

The given series is: ${{3 \times {1^3}} \over {{1^2}}} + {{5 \times ({1^3} + {2^3})} \over {{1^2} + {2^2}}} + {{7 \times \left( {{1^3} + {2^3} + {3^3}} \right)} \over {{1^2} + {2^2} + {3^2}}} + \dots$ We need to find the $r$-th term, $T_r$. Observing the pattern: The coefficient in the numerator is $3, 5, 7, \dots$, which is an arithmetic progression. The $r$-th term of this AP is $2r+1$. The sum in the numerator is $1^3 + 2^3 + \dots + r^3$, which is the sum of the cubes of the first $r$ natural numbers. The sum in the denominator is $1^2 + 2^2 + \dots + r^2$, which is the sum of the squares of the first $r$ natural numbers.

Therefore, the general term $T_r$ is: $T_r = \frac{(2r+1) \sum_{k=1}^r k^3}{\sum_{k=1}^r k^2}$

Step 2: Simplify the General Term ($T_r$)

Using the formulas for the sum of cubes and sum of squares: $\sum_{k=1}^r k^3 = \left(\frac{r(r+1)}{2}\right)^2 = \frac{r^2(r+1)^2}{4}$ $\sum_{k=1}^r k^2 = \frac{r(r+1)(2r+1)}{6}$ Substitute these into the expression for $T_r$: $T_r = \frac{(2r+1) \left(\frac{r^2(r+1)^2}{4}\right)}{\frac{r(r+1)(2r+1)}{6}}$ To simplify, multiply the numerator by the reciprocal of the denominator: $T_r = (2r+1) \times \frac{r^2(r+1)^2}{4} \times \frac{6}{r(r+1)(2r+1)}$ Cancel out common terms: $T_r = \frac{\cancel{(2r+1)} \cdot r^{\cancel{2}} \cdot (r+1)^{\cancel{2}} \cdot \cancel{6}^3}{\cancel{4}_2 \cdot \cancel{r} \cdot \cancel{(r+1)} \cdot \cancel{(2r+1)}}$ This simplifies to: $T_r = \frac{3r(r+1)}{2}$ Expand this expression for easier summation: $T_r = \frac{3}{2}(r^2 + r)$

Step 3: Calculate the Sum of the Series up to 10 Terms ($S_{10}$)

We need to find the sum of the first 10 terms, which is $S_{10} = \sum_{r=1}^{10} T_r$. $S_{10} = \sum_{r=1}^{10} \frac{3}{2}(r^2 + r)$ Factor out the constant $\frac{3}{2}$: $S_{10} = \frac{3}{2} \sum_{r=1}^{10} (r^2 + r)$ Separate the summation: $S_{10} = \frac{3}{2} \left( \sum_{r=1}^{10} r^2 + \sum_{r=1}^{10} r \right)$ Now, apply the sum formulas for $n=10$: $\sum_{r=1}^{10} r^2 = \frac{10(10+1)(2 \cdot 10+1)}{6} = \frac{10 \cdot 11 \cdot 21}{6} = \frac{2310}{6} = 385$ $\sum_{r=1}^{10} r = \frac{10(10+1)}{2} = \frac{10 \cdot 11}{2} = \frac{110}{2} = 55$ Substitute these values back into the expression for $S_{10}$: $S_{10} = \frac{3}{2} (385 + 55)$ $S_{10} = \frac{3}{2} (440)$ $S_{10} = 3 \times 220$ $S_{10} = 660$

Common Mistakes & Tips

• Formula Recall: Ensure accurate recall of the sum formulas for $k$, $k^2$, and $k^3$. Errors here will propagate.
• Algebraic Simplification: Be meticulous when simplifying the general term $T_r$. Canceling terms incorrectly is a common pitfall.
• Final Calculation: Double-check the arithmetic when substituting values into the sum formulas and performing the final multiplication.

Summary

The problem involves summing a series where each term is a ratio of sums of powers. We first identified the general term $T_r$ by observing the pattern in the numerator's coefficient and the sums of cubes and squares. Using standard formulas, we simplified $T_r$ to $\frac{3r(r+1)}{2}$. Finally, we computed the sum of the first 10 terms of this simplified series by applying the formulas for the sum of natural numbers and their squares, resulting in a sum of 660.

The final answer is \boxed{660}.