If `1^2+2^2+3^2+.....+x^2=((x)(x+1)(2x+1))/6`, then the value of `1^2+3^2+5^2+.......+19^2` is equal to:

Options

A. a)1330
B. b)2100
C. c)1485
D. d)2500

Answer

To solve the problem of finding the value of \(1^2 + 3^2 + 5^2 + \ldots + 19^2\), we can follow these steps:

### Step 1: Identify the series
The series \(1^2 + 3^2 + 5^2 + \ldots + 19^2\) consists of the squares of the first \(n\) odd natural numbers.

### Step 2: Determine the number of terms
The odd numbers from 1 to 19 are \(1, 3, 5, 7, 9, 11, 13, 15, 17, 19\). There are a total of 10 terms in this series.

### Step 3: Use the formula for the sum of squares of the first \(n\) odd natural numbers
The formula for the sum of squares of the first \(n\) odd natural numbers is given by:

\[ \text{Sum} = \frac{n}{3} \times (2n + 1) \times (2n - 1) \]

Where \(n\) is the number of odd terms.

### Step 4: Substitute \(n = 10\) into the formula
Now substituting \(n = 10\):

\[ \text{Sum} = \frac{10}{3} \times (2 \times 10 + 1) \times (2 \times 10 - 1) \]

### Step 5: Calculate the values
Calculating the terms inside the formula:

- \(2n + 1 = 2 \times 10 + 1 = 20 + 1 = 21\)
- \(2n - 1 = 2 \times 10 - 1 = 20 - 1 = 19\)

Now substituting these values back into the formula:

\[ \text{Sum} = \frac{10}{3} \times 21 \times 19 \]

### Step 6: Perform the multiplication
Calculating \(21 \times 19\):

\[ 21 \times 19 = 399 \]

Now substituting back:

\[ \text{Sum} = \frac{10}{3} \times 399 \]

### Step 7: Final calculation
Now, performing the division:

\[ \frac{3990}{3} = 1330 \]

### Final Answer
Thus, the value of \(1^2 + 3^2 + 5^2 + \ldots + 19^2\) is:

\[ \boxed{1330} \]