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

Options

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

Answer

To solve the expression \(1^2 - 2^2 + 3^2 - 4^2 + \ldots - 198^2 + 199^2\), we can group the terms in pairs and simplify using the difference of squares formula. Here’s a step-by-step solution:

### 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: Apply the difference of squares
Using the formula \(a^2 - b^2 = (a-b)(a+b)\), we can simplify each pair:
- For \(1^2 - 2^2\):
\[ 1^2 - 2^2 = (1 - 2)(1 + 2) = (-1)(3) = -3 \]
- For \(3^2 - 4^2\):
\[ 3^2 - 4^2 = (3 - 4)(3 + 4) = (-1)(7) = -7 \]
- For \(5^2 - 6^2\):
\[ 5^2 - 6^2 = (5 - 6)(5 + 6) = (-1)(11) = -11 \]
- Continuing this pattern, we see that each pair contributes a negative odd number.

### Step 3: Identify the pattern
The pairs give us:
\[ -3, -7, -11, -15, \ldots, -395 \]
This is an arithmetic sequence where the first term \(a = -3\) and the common difference \(d = -4\).

### Step 4: Find the number of terms
To find the last term, we set:
\[ -3 - 4(n-1) = -395 \]
Solving for \(n\):
\[ -4(n-1) = -395 + 3 \\ -4(n-1) = -392 \\ n-1 = 98 \\ n = 99 \]
So there are 99 pairs.

### Step 5: Sum the arithmetic series
The sum \(S\) of the first \(n\) terms of an arithmetic series can be calculated using:
\[ S_n = \frac{n}{2} (a + l) \]
where \(l\) is the last term. Here:
\[ S = \frac{99}{2} (-3 - 395) = \frac{99}{2} \times (-398) = -99 \times 199 = -19701 \]

### Step 6: Add the last term
Now, we need to add \(199^2\):
\[ 199^2 = 39601 \]
So the final result is:
\[ -19701 + 39601 = 19900 \]

### Final Answer
Thus, the value of the expression \(1^2 - 2^2 + 3^2 - 4^2 + \ldots - 198^2 + 199^2\) is:
\[ \boxed{19900} \]