`1^2 - 2^2 + 3^2 - 4^2 + ….. - 198^2 + 199^2 ` :

Options

A. 19900
B. 12321
C. 19998
D. none of these

Answer

To solve the expression \(1^2 - 2^2 + 3^2 - 4^2 + \ldots - 198^2 + 199^2\), we can follow these steps:

### Step 1: Group the terms
We can group the terms in pairs:
\[ (1^2 - 2^2) + (3^2 - 4^2) + (5^2 - 6^2) + \ldots + (197^2 - 198^2) + 199^2 \]

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

### Step 3: Write the general form for the pairs
The general form for the pairs can be written as:
\[ -(4n - 1) \]
where \(n\) takes values from 1 to 99 (since there are 198 terms from 1 to 198).

### Step 4: Calculate the sum of the pairs
The total number of pairs is 99 (from 1 to 198). The sum of the pairs is:
\[ \sum_{n=1}^{99} -(4n - 1) = -\left(4\sum_{n=1}^{99} n - \sum_{n=1}^{99} 1\right) \]
The sum of the first 99 natural numbers is:
\[ \sum_{n=1}^{99} n = \frac{99 \times 100}{2} = 4950 \]
And the sum of 1 for 99 terms is:
\[ \sum_{n=1}^{99} 1 = 99 \]
Thus, we have:
\[ - \left(4 \times 4950 - 99\right) = - (19800 - 99) = -19701 \]

### Step 5: Add the last term
Now, we add the last term \(199^2\):
\[ 199^2 = 39601 \]
So, the final result is:
\[ -19701 + 39601 = 19900 \]

### Final Answer
The value of the expression \(1^2 - 2^2 + 3^2 - 4^2 + \ldots - 198^2 + 199^2\) is \(19900\).

---