The sum of the series 1-3+5-7+9-11+ . . . . To n terms is

Options

A. `-n, " when n is even "G373`
B. 2n, when n is even
C. `-n," when n is odd"
D. 2n, when n is odd

Answer

To find the sum of the series \(1 - 3 + 5 - 7 + 9 - 11 + \ldots\) up to \(n\) terms, we can follow these steps:

### Step 1: Identify the Pattern
The series alternates between positive and negative terms. The positive terms are the odd numbers starting from 1 (i.e., \(1, 5, 9, \ldots\)) and the negative terms are also odd numbers (i.e., \(3, 7, 11, \ldots\)).

### Step 2: Group the Terms
We can group the terms in pairs:
\[ (1 - 3) + (5 - 7) + (9 - 11) + \ldots \]
Each pair simplifies to:
\[ 1 - 3 = -2, \quad 5 - 7 = -2, \quad 9 - 11 = -2 \]
Thus, each pair contributes \(-2\) to the sum.

### Step 3: Determine the Number of Pairs
If \(n\) is even, we can form \(\frac{n}{2}\) pairs. If \(n\) is odd, the last term will be unpaired. However, for the sake of this solution, we will consider \(n\) to be even.

### Step 4: Calculate the Sum
The total sum \(S_n\) when \(n\) is even can be calculated as:
\[ S_n = \text{Number of pairs} \times \text{Sum of each pair} = \frac{n}{2} \times (-2) \]
This simplifies to:
\[ S_n = -n \]

### Conclusion
Thus, the sum of the series \(1 - 3 + 5 - 7 + 9 - 11 + \ldots\) up to \(n\) terms (where \(n\) is even) is:
\[ S_n = -n \]