The sum of n terms of the series , where n is an even number
`1^2 - 2^2 + 3^2 - 4^(2) + 5^(2) - 6^2 + …` :

Options

A. `n(n+1)`
B. `(n(n+1))/(2)`
C. `-(n(n+1))/(2)`
D. none of these

Answer

To find the sum of the first n terms of the series \(1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + \ldots\) where n is an even number, we can follow these steps:

### Step 1: Identify the pattern in the series
The series alternates between positive and negative squares of integers. We can group the terms in pairs:
\[ (1^2 - 2^2) + (3^2 - 4^2) + (5^2 - 6^2) + \ldots \]

### Step 2: Simplify each pair
Each pair can be simplified using the difference of squares formula:
\[ a^2 - b^2 = (a-b)(a+b) \]
For the first pair:
\[ 1^2 - 2^2 = (1 - 2)(1 + 2) = (-1)(3) = -3 \]
For the second pair:
\[ 3^2 - 4^2 = (3 - 4)(3 + 4) = (-1)(7) = -7 \]
For the third pair:
\[ 5^2 - 6^2 = (5 - 6)(5 + 6) = (-1)(11) = -11 \]
Thus, the sum of each pair is:
\[ -3, -7, -11, \ldots \]

### Step 3: Generalize the sum of the pairs
The k-th pair can be represented as:
\[ (2k-1)^2 - (2k)^2 = -((2k-1) + 2k)((2k-1) - 2k) = -((4k-1)(-1)) = 4k - 1 \]
So, the sum of the first k pairs is:
\[ \text{Sum} = -3 - 7 - 11 - \ldots - (4k - 1) \]

### Step 4: Calculate the total number of pairs
Since n is even, the number of pairs is \(k = \frac{n}{2}\).

### Step 5: Sum of the first k terms of the sequence
The sum of the first k terms of the sequence \(-3, -7, -11, \ldots\) can be calculated as:
\[ \text{Sum} = -\left(3 + 7 + 11 + \ldots + (4k - 1)\right) \]
This is an arithmetic series where:
- First term \(a = 3\)
- Common difference \(d = 4\)
- Number of terms \(k\)

The sum of the first k terms of an arithmetic series is given by:
\[ S_k = \frac{k}{2} \times (2a + (k-1)d) \]
Substituting the values:
\[ S_k = \frac{k}{2} \times (2 \cdot 3 + (k-1) \cdot 4) = \frac{k}{2} \times (6 + 4k - 4) = \frac{k}{2} \times (4k + 2) = k(2k + 1) \]

### Step 6: Substitute back to find the total sum
Thus, the total sum becomes:
\[ \text{Total Sum} = -k(2k + 1) \]
Substituting \(k = \frac{n}{2}\):
\[ \text{Total Sum} = -\frac{n}{2}(2 \cdot \frac{n}{2} + 1) = -\frac{n}{2}(n + 1) = -\frac{n(n + 1)}{2} \]

### Final Result
The sum of the first n terms of the series is:
\[ \text{Sum} = -\frac{n(n + 1)}{2} \]