Key Concepts and Formulas

• General Term ($T_n$) Identification: The ability to discern a pattern and express the $n$-th term of a series using a formula is crucial.
• Summation Formulas: Standard formulas for the sum of powers of the first $n$ natural numbers are essential:
  • $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$
  • $\sum_{k=1}^n k^3 = \left(\frac{n(n+1)}{2}\right)^2$
• Arithmetic Progression (AP): Recognizing and using the formula for the $n$-th term of an AP ($a_n = a + (n-1)d$).

Step-by-Step Solution

Step 1: Analyze the Structure of the Series and Identify Patterns

The given series is: $S = 1 + 6 + \frac{9(1^2 + 2^2 + 3^2)}{7} + \frac{12(1^2 + 2^2 + 3^2 + 4^2)}{9} + \frac{15(1^2 + 2^2 + \dots + 5^2)}{11} + \dots$ Let's examine the terms starting from the third term to find a general pattern. Let $T_n$ be the $n$-th term of the series.

• Numerator Coefficient: The coefficients are $9, 12, 15, \dots$. This is an arithmetic progression (AP) with the first term $a=9$ and common difference $d=3$. For the $n$-th term of the series, the corresponding term in this AP (which starts from the 3rd term of the series) is the $(n-2)$-th term of the AP. So, the coefficient is $9 + ((n-2)-1)3 = 9 + (n-3)3 = 9 + 3n - 9 = 3n$. This formula is valid for $n \ge 3$.
• Sum of Squares Part: The series involves sums of squares. For the $n$-th term of the series (where $n \ge 3$), the sum of squares goes up to $n^2$. This is represented by $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$.
• Denominator: The denominators are $7, 9, 11, \dots$. This is an AP with the first term $a=7$ and common difference $d=2$. For the $n$-th term of the series (where $n \ge 3$), the corresponding term in this AP is the $(n-2)$-th term of the AP. So, the denominator is $7 + ((n-2)-1)2 = 7 + (n-3)2 = 7 + 2n - 6 = 2n+1$. This formula is valid for $n \ge 3$.

Combining these observations for $n \ge 3$, the general term $T_n$ is: $T_n = \frac{(3n) \left( \sum_{k=1}^n k^2 \right)}{2n+1}$ Substituting the formula for the sum of squares: $T_n = \frac{3n \cdot \frac{n(n+1)(2n+1)}{6}}{2n+1}$ We can cancel out the $(2n+1)$ term and simplify the constants: $T_n = \frac{3n \cdot n(n+1)}{6} = \frac{n^2(n+1)}{2}$

Step 2: Verify the General Term for the Initial Terms

We found $T_n = \frac{n^2(n+1)}{2}$ by analyzing terms from the third onwards. Let's check if this formula holds for the first two terms:

• For $n=1$: $T_1 = \frac{1^2(1+1)}{2} = \frac{1 \cdot 2}{2} = 1$. This matches the first term.
• For $n=2$: $T_2 = \frac{2^2(2+1)}{2} = \frac{4 \cdot 3}{2} = 6$. This matches the second term.

Since the formula $T_n = \frac{n^2(n+1)}{2}$ is valid for $n=1, 2, 3, \dots$, it is the correct general term for the entire series.

Step 3: Express the General Term in a Summation-Friendly Form

To facilitate summation, we expand the general term: $T_n = \frac{n^2(n+1)}{2} = \frac{n^3 + n^2}{2} = \frac{1}{2}(n^3 + n^2)$

Step 4: Calculate the Sum of the Series up to 15 Terms

We need to find the sum of the first 15 terms, $S_{15} = \sum_{n=1}^{15} T_n$. $S_{15} = \sum_{n=1}^{15} \frac{1}{2}(n^3 + n^2)$ We can pull out the constant $\frac{1}{2}$ and split the summation: $S_{15} = \frac{1}{2} \left( \sum_{n=1}^{15} n^3 + \sum_{n=1}^{15} n^2 \right)$ Now, we use the standard summation formulas with $n=15$:

• Sum of cubes: $\sum_{n=1}^{15} n^3 = \left(\frac{15(15+1)}{2}\right)^2 = \left(\frac{15 \cdot 16}{2}\right)^2 = (15 \cdot 8)^2 = (120)^2 = 14400$
• Sum of squares: $\sum_{n=1}^{15} n^2 = \frac{15(15+1)(2 \cdot 15+1)}{6} = \frac{15 \cdot 16 \cdot 31}{6} = \frac{240 \cdot 31}{6} = 40 \cdot 31 = 1240$

Step 5: Final Calculation

Substitute the calculated sums back into the expression for $S_{15}$: $S_{15} = \frac{1}{2} (14400 + 1240)$ $S_{15} = \frac{1}{2} (15640)$ $S_{15} = 7820$

This result matches option (D).

Common Mistakes & Tips

• Indexing Errors: Be extremely careful when determining the correct index for the arithmetic progressions that form parts of the general term. Ensure the formula for the $n$-th term of the series correctly corresponds to the terms in the APs.
• Algebraic Simplification: Mistakes can arise during the simplification of the general term, especially when canceling terms like $(2n+1)$. It's advisable to write out the intermediate steps clearly.
• Arithmetic Errors: Double-check all calculations, particularly when dealing with squares and products of larger numbers, to avoid errors in the final sum.

Summary

The problem requires identifying the general term of a complex series by analyzing the patterns in its numerator coefficients, sum of squares components, and denominators. After establishing the general term $T_n = \frac{n^2(n+1)}{2}$ and verifying its validity for all terms, the sum of the series up to 15 terms is computed by applying standard formulas for the sum of cubes and squares. The calculation yields a sum of 7820.

The final answer is $\boxed{7820}$, which corresponds to option (D).