The value of `1^(2) + 3^(2) + 5^(2) + ..... + 25^(2)` is

Options

A. 1728
B. 1456
C. 2925
D. 1469

Answer

To find the value of \(1^2 + 3^2 + 5^2 + \ldots + 25^2\), we can follow these steps:

### Step 1: Identify the sequence
The numbers in the series are the squares of the first 13 odd numbers. The odd numbers can be expressed in the form \(2n - 1\), where \(n\) is a positive integer.

### Step 2: Determine the last term
The last term in our series is \(25\). We need to find \(n\) such that:
\[ 2n - 1 = 25 \]
Solving for \(n\):
\[ 2n = 26 \implies n = 13 \]
Thus, we will sum the squares of the first 13 odd numbers.

### Step 3: Write the summation
We can express the sum as:
\[ \sum_{n=1}^{13} (2n - 1)^2 \]

### Step 4: Expand the square
Expanding \((2n - 1)^2\):
\[ (2n - 1)^2 = 4n^2 - 4n + 1 \]
So, we can rewrite the summation:
\[ \sum_{n=1}^{13} (2n - 1)^2 = \sum_{n=1}^{13} (4n^2 - 4n + 1) \]

### Step 5: Split the summation
We can separate the summation:
\[ \sum_{n=1}^{13} (4n^2 - 4n + 1) = 4\sum_{n=1}^{13} n^2 - 4\sum_{n=1}^{13} n + \sum_{n=1}^{13} 1 \]

### Step 6: Calculate each summation
1. **Sum of the first \(n\) natural numbers**:
\[ \sum_{n=1}^{k} n = \frac{k(k + 1)}{2} \]
For \(k = 13\):
\[ \sum_{n=1}^{13} n = \frac{13 \times 14}{2} = 91 \]

2. **Sum of the squares of the first \(n\) natural numbers**:
\[ \sum_{n=1}^{k} n^2 = \frac{k(k + 1)(2k + 1)}{6} \]
For \(k = 13\):
\[ \sum_{n=1}^{13} n^2 = \frac{13 \times 14 \times 27}{6} = 819 \]

3. **Sum of 1's**:
\[ \sum_{n=1}^{13} 1 = 13 \]

### Step 7: Substitute back into the equation
Now substituting these values back:
\[ 4 \sum_{n=1}^{13} n^2 = 4 \times 819 = 3276 \]
\[ -4 \sum_{n=1}^{13} n = -4 \times 91 = -364 \]
\[ \sum_{n=1}^{13} 1 = 13 \]

Combining these:
\[ 3276 - 364 + 13 = 2925 \]

### Final Answer
Thus, the value of \(1^2 + 3^2 + 5^2 + \ldots + 25^2\) is:
\[ \boxed{2925} \]