The sum of series
`1+6+(9(1^(2)+2^(2)+3^(2)))/(7) + (12(1^(2)+2^(2)+3^(2)+4^(2)))/(9)+(15(1^(2)+2^(2)+...+5^(2)))/(11)+...` up to 15 terms is

Options

A. 7820
B. 7830
C. 7520
D. 7510

Answer

To find the sum of the series given in the question, we will analyze the series term by term and derive a general formula for the nth term.

### Step-by-Step Solution:

1. **Identify the Series Pattern**:
The series starts with:
\[ S = 1 + 6 + \frac{9(1^2 + 2^2 + 3^2)}{7} + \frac{12(1^2 + 2^2 + 3^2 + 4^2)}{9} + \frac{15(1^2 + 2^2 + 3^2 + 4^2 + 5^2)}{11} + \ldots \]
We can see that the series has a pattern where each term has a specific structure.

2. **Generalize the nth Term**:
The nth term can be expressed as:
\[ T_n = \frac{(3n + 3)(1^2 + 2^2 + \ldots + n^2)}{(2n + 5)} \]
where \(3n + 3\) represents the coefficients (6, 9, 12, 15, ...) and \(2n + 5\) represents the denominators (7, 9, 11, ...).

3. **Sum of Squares Formula**:
The sum of squares of the first n natural numbers is given by:
\[ 1^2 + 2^2 + \ldots + n^2 = \frac{n(n + 1)(2n + 1)}{6} \]

4. **Substituting the Sum of Squares**:
Substitute the sum of squares into the nth term:
\[ T_n = \frac{(3n + 3) \cdot \frac{n(n + 1)(2n + 1)}{6}}{(2n + 5)} \]

5. **Simplifying the nth Term**:
Simplifying gives:
\[ T_n = \frac{(3n + 3) \cdot n(n + 1)(2n + 1)}{6(2n + 5)} \]

6. **Finding the Total Sum**:
Now, we need to sum \(T_n\) from \(n = 1\) to \(n = 15\):
\[ S = \sum_{n=1}^{15} T_n \]

7. **Calculating the Sum**:
We can compute this sum directly or use properties of summations. The sum can be calculated by substituting values from 1 to 15 into the expression for \(T_n\).

8. **Final Calculation**:
After calculating each term and summing them up, we find:
\[ S = 1 + 6 + \sum_{n=1}^{15} \frac{(3n + 3) \cdot \frac{n(n + 1)(2n + 1)}{6}}{(2n + 5)} \]