What is the sum of the first 50 terms of the series `(1xx3)+(3xx5)+(5xx7)+………….?`

Options

A. 1,71,650
B. 26600
C. 26650
D. 26900

Explanation

To find the sum of the first 50 terms of the series \( (1 \times 3) + (3 \times 5) + (5 \times 7) + \ldots \), we can follow these steps:

### Step 1: Identify the nth term of the series
The series can be expressed in terms of \( n \):
- The first term is \( 1 \times 3 \)
- The second term is \( 3 \times 5 \)
- The third term is \( 5 \times 7 \)

The pattern shows that the nth term can be represented as:
\[ T_n = (2n - 1)(2n + 1) \]
This is because:
- The first factor \( 2n - 1 \) gives the odd numbers starting from 1 (1, 3, 5, ...).
- The second factor \( 2n + 1 \) gives the next odd number after \( 2n - 1 \).

### Step 2: Simplify the nth term
Now, let's simplify \( T_n \):
\[ T_n = (2n - 1)(2n + 1) = 4n^2 - 1 \]

### Step 3: Find the sum of the first 50 terms
We need to find the sum \( S_n \) of the first \( n \) terms:
\[ S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (4k^2 - 1) \]
This can be separated into two sums:
\[ S_n = 4\sum_{k=1}^{n} k^2 - \sum_{k=1}^{n} 1 \]
The sum of the first \( n \) natural numbers squared is given by:
\[ \sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6} \]
And the sum of 1 for \( n \) terms is simply \( n \).

### Step 4: Substitute and calculate \( S_{50} \)
Substituting \( n = 50 \):
\[ S_{50} = 4 \left(\frac{50(50 + 1)(2 \cdot 50 + 1)}{6}\right) - 50 \]
Calculating \( \sum_{k=1}^{50} k^2 \):
\[ \sum_{k=1}^{50} k^2 = \frac{50 \cdot 51 \cdot 101}{6} = \frac{257550}{6} = 42925 \]
Now substituting back into \( S_{50} \):
\[ S_{50} = 4 \cdot 42925 - 50 = 171700 - 50 = 171650 \]

### Final Answer
Thus, the sum of the first 50 terms of the series is:
\[ \boxed{171650} \]