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 sum of the squares of the even numbers from 2 to 100, and subtract the sum of the squares of the odd numbers from 1 to 99.

Let's break it down step by step:

### Step 1: Identify the series
The even numbers from 2 to 100 can be expressed as:
\[ 2, 4, 6, \ldots, 100 \]
The odd numbers from 1 to 99 can be expressed as:
\[ 1, 3, 5, \ldots, 99 \]

### Step 2: Write the sums
We need to find:
\[ S = (2^2 + 4^2 + 6^2 + \ldots + 100^2) - (1^2 + 3^2 + 5^2 + \ldots + 99^2) \]

### Step 3: Pair the terms
We can pair the terms in the following way:
\[ (2^2 - 1^2) + (4^2 - 3^2) + (6^2 - 5^2) + \ldots + (100^2 - 99^2) \]

### Step 4: Use the difference of squares
Recall the identity:
\[ a^2 - b^2 = (a - b)(a + b) \]
Using this, we can rewrite each pair:
- For \( 2^2 - 1^2 \):
\[ (2 - 1)(2 + 1) = 1 \cdot 3 = 3 \]
- For \( 4^2 - 3^2 \):
\[ (4 - 3)(4 + 3) = 1 \cdot 7 = 7 \]
- For \( 6^2 - 5^2 \):
\[ (6 - 5)(6 + 5) = 1 \cdot 11 = 11 \]
- Continuing this pattern, we see that:
\[ (2n)^2 - (2n-1)^2 = (2n - (2n-1))(2n + (2n-1)) = 1 \cdot (4n - 1) = 4n - 1 \]

### Step 5: Determine the number of pairs
The even numbers from 2 to 100 are:
\[ 2, 4, 6, \ldots, 100 \]
This is an arithmetic series with:
- First term \( a = 2 \)
- Last term \( l = 100 \)
- Common difference \( d = 2 \)

The number of terms \( n \) can be calculated as:
\[ n = \frac{l - a}{d} + 1 = \frac{100 - 2}{2} + 1 = 50 \]

### Step 6: Calculate the total sum
The sum of the differences is:
\[ S = (3 + 7 + 11 + \ldots + (4n - 1)) \]
This is an arithmetic series where:
- First term \( a = 3 \)
- Last term \( l = 4(50) - 1 = 199 \)
- Number of terms \( n = 50 \)

The sum of an arithmetic series is given by:
\[ S_n = \frac{n}{2} (a + l) \]
Substituting the values:
\[ S = \frac{50}{2} (3 + 199) = 25 \cdot 202 = 5050 \]

### Final Answer
Thus, the result of the expression is:
\[ \boxed{5050} \]