The sum of :
`(2^2 + 4^2 + 6^2 + …….+ 100^2) - (1^2 + 3^2 + 5^2 + …+ 99^2)` is :

Options

A. 5555
B. 5050
C. 888
D. 222

Answer

To solve the problem, we need to calculate the difference between the sum of the squares of the even numbers from 2 to 100 and the sum of the squares of the odd numbers from 1 to 99.

### Step-by-Step Solution:

1. **Identify the two series**:
- Even numbers: \(2^2 + 4^2 + 6^2 + \ldots + 100^2\)
- Odd numbers: \(1^2 + 3^2 + 5^2 + \ldots + 99^2\)

2. **Express the sums**:
- The sum of the squares of the even numbers can be expressed as:
\[ \sum_{k=1}^{50} (2k)^2 = 4 \sum_{k=1}^{50} k^2 \]
- The sum of the squares of the odd numbers can be expressed as:
\[ \sum_{k=0}^{49} (2k + 1)^2 = \sum_{k=0}^{49} (4k^2 + 4k + 1) = 4 \sum_{k=0}^{49} k^2 + 4 \sum_{k=0}^{49} k + \sum_{k=0}^{49} 1 \]

3. **Calculate the sums**:
- The formula for the sum of the first \(n\) squares is:
\[ \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \]
- For \(n = 50\):
\[ \sum_{k=1}^{50} k^2 = \frac{50 \times 51 \times 101}{6} = 42925 \]
- Therefore, the sum of the squares of the even numbers:
\[ 4 \sum_{k=1}^{50} k^2 = 4 \times 42925 = 171700 \]

4. **Calculate the odd numbers sum**:
- For \(n = 49\):
\[ \sum_{k=0}^{49} k^2 = \frac{49 \times 50 \times 99}{6} = 40425 \]
- The sum of the first 49 natural numbers:
\[ \sum_{k=0}^{49} k = \frac{49 \times 50}{2} = 1225 \]
- The number of terms is 50 (from 0 to 49):
\[ \sum_{k=0}^{49} 1 = 50 \]
- Therefore, the sum of the squares of the odd numbers:
\[ 4 \times 40425 + 4 \times 1225 + 50 = 161700 + 4900 + 50 = 166650 \]

5. **Calculate the final difference**:
- Now we subtract the sum of the odd squares from the sum of the even squares:
\[ (2^2 + 4^2 + 6^2 + \ldots + 100^2) - (1^2 + 3^2 + 5^2 + \ldots + 99^2) = 171700 - 166650 = 5050 \]

### Final Answer:
The result of the expression is \(5050\).