Key Concepts and Formulas

1. Sum of the first $n$ cubes: $\sum_{k=1}^n k^3 = \left(\frac{n(n+1)}{2}\right)^2$
2. Sum of the first $n$ odd natural numbers: $\sum_{k=1}^n (2k-1) = n^2$
3. Sum of the first $n$ natural numbers: $\sum_{k=1}^n k = \frac{n(n+1)}{2}$
4. Sum of the squares of the first $n$ natural numbers: $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$
5. Linearity of Summation: $\sum_{k=1}^n (a A_k + b B_k) = a \sum_{k=1}^n A_k + b \sum_{k=1}^n B_k$

Step-by-Step Solution

Step 1: Determine and Simplify the General Term ($T_n$) The given series is ${{{1^3}} \over 1} + {{{1^3} + {2^3}} \over {1 + 3}} + {{{1^3} + {2^3} + {3^3}} \over {1 + 3 + 5}} + \dots$ We need to find the $n^{th}$ term ($T_n$) of this series.

• Numerator of $T_n$: The numerator of the $n^{th}$ term is the sum of the cubes of the first $n$ natural numbers. $N_n = 1^3 + 2^3 + 3^3 + \dots + n^3 = \sum_{k=1}^n k^3$ Using the formula for the sum of the first $n$ cubes: $N_n = \left(\frac{n(n+1)}{2}\right)^2$ Why: This formula directly represents the sum of cubes as seen in the numerator's pattern.

• Denominator of $T_n$: The denominator of the $n^{th}$ term is the sum of the first $n$ odd natural numbers. $D_n = 1 + 3 + 5 + \dots + (2n-1) = \sum_{k=1}^n (2k-1)$ Using the formula for the sum of the first $n$ odd numbers: $D_n = n^2$ Why: This formula represents the sum of consecutive odd numbers, matching the pattern in the denominator.

• Formulate and Simplify $T_n$: The $n^{th}$ term $T_n$ is the ratio of its numerator and denominator: $T_n = \frac{N_n}{D_n} = \frac{\left(\frac{n(n+1)}{2}\right)^2}{n^2}$ Now, we simplify this expression: $T_n = \frac{\frac{n^2(n+1)^2}{4}}{n^2}$ Since $n \ge 1$ for terms in a series, $n^2 \neq 0$, so we can cancel $n^2$: $T_n = \frac{(n+1)^2}{4}$ Expanding the term $(n+1)^2$: $T_n = \frac{n^2 + 2n + 1}{4} = \frac{1}{4}(n^2 + 2n + 1)$ Why: Simplifying $T_n$ into a polynomial form makes it amenable to standard summation techniques.

Step 2: Calculate the Sum of the First 9 Terms ($S_9$) We need to find the sum of the first 9 terms, $S_9 = \sum_{n=1}^9 T_n$. $S_9 = \sum_{n=1}^9 \frac{1}{4}(n^2 + 2n + 1)$ Why: This is the summation of the simplified general term, which is the objective of the problem.

• Apply Linearity of Summation: We can pull out the constant factor $\frac{1}{4}$ and distribute the summation to each term inside the parenthesis: $S_9 = \frac{1}{4} \left( \sum_{n=1}^9 n^2 + \sum_{n=1}^9 2n + \sum_{n=1}^9 1 \right)$ $S_9 = \frac{1}{4} \left( \sum_{n=1}^9 n^2 + 2 \sum_{n=1}^9 n + \sum_{n=1}^9 1 \right)$ Why: The linearity property allows us to break down the summation of a polynomial into sums of powers of $n$ and constants, which have known formulas.

• Apply Standard Summation Formulas for $N=9$: Now, we evaluate each summation using the standard formulas with $n=9$:

  1. Sum of squares: $\sum_{n=1}^9 n^2 = \frac{9(9+1)(2 \times 9 + 1)}{6} = \frac{9 \times 10 \times 19}{6} = \frac{1710}{6} = 285$
  2. Sum of natural numbers: $\sum_{n=1}^9 n = \frac{9(9+1)}{2} = \frac{9 \times 10}{2} = \frac{90}{2} = 45$
  3. Sum of a constant: $\sum_{n=1}^9 1 = 9$ Why: These are direct applications of the fundamental summation formulas for $n^2$, $n$, and a constant.

• Substitute and Calculate the Final Sum: Substitute the calculated values back into the expression for $S_9$: $S_9 = \frac{1}{4} (285 + 2(45) + 9)$ $S_9 = \frac{1}{4} (285 + 90 + 9)$ $S_9 = \frac{1}{4} (384)$ $S_9 = 96$ Why: This final calculation combines all the evaluated parts to arrive at the total sum of the first 9 terms.

Common Mistakes & Tips

• Incorrect General Term: Always verify your derived $T_n$ by plugging in $n=1, 2, 3$ to ensure it matches the given series terms.
• Algebraic Errors: Be extremely careful with algebraic manipulations, especially when expanding squares and simplifying fractions in $T_n$.
• Formula Recall: Ensure you have memorized the standard summation formulas for $k$, $k^2$, $k^3$, and sums of arithmetic/geometric progressions, as they are frequently tested.

Summary To find the sum of the first 9 terms of the given series, we first identified the general term ($T_n$) by analyzing the patterns in the numerator (sum of cubes) and the denominator (sum of odd numbers). We then simplified $T_n$ to a polynomial form using standard summation formulas. Finally, we applied the linearity of summation and the formulas for the sum of squares, sum of natural numbers, and sum of a constant to calculate the sum of the first 9 terms.

The final answer is $\boxed{96}$, which corresponds to option (D).