Find the sum of n-terms of the series `2.5 + 5.8 + 8.11 +……..`

Answer

To find the sum of the n-terms of the series \(2.5 + 5.8 + 8.11 + \ldots\), we need to analyze the pattern of the terms in the series.

### Step-by-step Solution:

1. **Identify the Terms**:
The series can be expressed as:
\[ 2 \times 5 + 5 \times 8 + 8 \times 11 + \ldots \]
Let's denote the terms as \(a_n = a_{n-1} + 3\), where the first term \(a_1 = 2\) and the second term \(b_n = b_{n-1} + 3\), where the first term \(b_1 = 5\).

2. **Find the nth Term**:
The first sequence \(a_n\) is an arithmetic progression (AP) with:
- First term \(a = 2\)
- Common difference \(d = 3\)

The nth term of this AP is given by:
\[ a_n = a + (n-1)d = 2 + (n-1) \cdot 3 = 3n - 1 \]

The second sequence \(b_n\) is also an AP with:
- First term \(b = 5\)
- Common difference \(d = 3\)

The nth term of this AP is given by:
\[ b_n = b + (n-1)d = 5 + (n-1) \cdot 3 = 3n + 2 \]

3. **Formulate the nth Term of the Series**:
The nth term of the series \(S_n\) can be expressed as:
\[ S_n = a_n \cdot b_n = (3n - 1)(3n + 2) \]

4. **Expand the nth Term**:
Expanding the product:
\[ S_n = (3n - 1)(3n + 2) = 9n^2 + 6n - 3n - 2 = 9n^2 + 3n - 2 \]

5. **Sum of n Terms**:
The sum of the first n terms \(S_n\) can be calculated as:
\[ S_n = \sum_{k=1}^{n} (9k^2 + 3k - 2) \]
This can be separated into three summations:
\[ S_n = 9 \sum_{k=1}^{n} k^2 + 3 \sum_{k=1}^{n} k - 2 \sum_{k=1}^{n} 1 \]

6. **Use Summation Formulas**:
- The formula for the sum of the first n squares is:
\[ \sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6} \]
- The formula for the sum of the first n natural numbers is:
\[ \sum_{k=1}^{n} k = \frac{n(n + 1)}{2} \]
- The sum of 1 for n terms is simply \(n\).

Plugging these into the equation:
\[ S_n = 9 \cdot \frac{n(n + 1)(2n + 1)}{6} + 3 \cdot \frac{n(n + 1)}{2} - 2n \]

7. **Simplify**:
\[ S_n = \frac{3}{2} n(n + 1)(2n + 1) + \frac{3}{2} n(n + 1) - 2n \]
Factor out \(\frac{3}{2} n(n + 1)\):
\[ S_n = \frac{3}{2} n(n + 1) \left(2n + 1 + 1\right) - 2n \]
This simplifies to:
\[ S_n = \frac{3}{2} n(n + 1)(2n + 2) - 2n \]

8. **Final Expression**:
After further simplification, we arrive at:
\[ S_n = 3n^3 + 6n^2 + n \]

### Final Answer:
The sum of the first n terms of the series is:
\[ S_n = 3n^3 + 6n^2 + n \]