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

\[ S = 1 + 6 + \frac{9(1^2 + 2^2 + 3^2)}{7} + \frac{12(1^2 + 2^2 + 3^2 + 4^2)}{9} + \frac{15(1^2 + 2^2 + 3^2 + 4^2 + 5^2)}{11} + \ldots \]

up to 15 terms, we first need to identify the pattern in the series.

### Step 1: Identify the pattern in the series

1. The first term is \(1\).
2. The second term is \(6\).
3. The third term is \(\frac{9(1^2 + 2^2 + 3^2)}{7}\).
4. The fourth term is \(\frac{12(1^2 + 2^2 + 3^2 + 4^2)}{9}\).
5. The fifth term is \(\frac{15(1^2 + 2^2 + 3^2 + 4^2 + 5^2)}{11}\).

From this, we can see that:

- The numerators follow the pattern \(3n\) for \(n = 1, 2, 3, \ldots\).
- The denominators follow the pattern \(2n + 1\) starting from \(3\).
- The sums in the numerators are the sums of squares of the first \(n\) natural numbers.

### Step 2: Write the nth term

The nth term can be expressed as:

\[ T_n = \frac{(3n)(1^2 + 2^2 + \ldots + n^2)}{(2n + 1)} \]

Using the formula for the sum of the squares of the first \(n\) natural numbers:

\[ 1^2 + 2^2 + \ldots + n^2 = \frac{n(n + 1)(2n + 1)}{6} \]

We can substitute this into our expression for \(T_n\):

\[ T_n = \frac{3n \cdot \frac{n(n + 1)(2n + 1)}{6}}{(2n + 1)} = \frac{n(n + 1)(2n + 1)}{4} \]

### Step 3: Find the sum of the first 15 terms

Now, we need to find the sum \(S_{15} = \sum_{n=1}^{15} T_n\):

\[ S_{15} = \sum_{n=1}^{15} \frac{n(n + 1)(2n + 1)}{4} \]

This can be simplified as:

\[ S_{15} = \frac{1}{4} \sum_{n=1}^{15} n(n + 1)(2n + 1) \]

### Step 4: Calculate the sum

To compute \(\sum_{n=1}^{15} n(n + 1)(2n + 1)\), we can expand it:

\[ n(n + 1)(2n + 1) = 2n^3 + 3n^2 + n \]

Thus, we need to calculate:

\[ \sum_{n=1}^{15} (2n^3 + 3n^2 + n) = 2\sum_{n=1}^{15} n^3 + 3\sum_{n=1}^{15} n^2 + \sum_{n=1}^{15} n \]

Using the formulas:

- \(\sum_{n=1}^{k} n = \frac{k(k + 1)}{2}\)
- \(\sum_{n=1}^{k} n^2 = \frac{k(k + 1)(2k + 1)}{6}\)
- \(\sum_{n=1}^{k} n^3 = \left(\frac{k(k + 1)}{2}\right)^2\)

For \(k = 15\):

1. \(\sum_{n=1}^{15} n = \frac{15 \cdot 16}{2} = 120\)
2. \(\sum_{n=1}^{15} n^2 = \frac{15 \cdot 16 \cdot 31}{6} = 1240\)
3. \(\sum_{n=1}^{15} n^3 = \left(\frac{15 \cdot 16}{2}\right)^2 = 14400\)

Now substituting these values:

\[ S_{15} = \frac{1}{4} \left(2 \cdot 14400 + 3 \cdot 1240 + 120\right) \]

Calculating:

\[ = \frac{1}{4} \left(28800 + 3720 + 120\right) = \frac{1}{4} \cdot 32740 = 8185 \]

### Final Answer

Thus, the sum of the series up to 15 terms is:

\[ \boxed{8185} \]