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 \( S_n = 1 - 3 + 5 - 7 + 9 - 11 + \ldots \) up to \( n \) terms, we can follow these steps:

### Step 1: Identify the pattern of the series
The series alternates between positive and negative odd numbers. The general term can be expressed as:
- For odd indexed terms (1st, 3rd, 5th, ...), the terms are positive: \( 1, 5, 9, \ldots \)
- For even indexed terms (2nd, 4th, 6th, ...), the terms are negative: \( -3, -7, -11, \ldots \)

### Step 2: Write the series in a more manageable form
We can express the series in terms of \( n \):
- If \( n \) is even, say \( n = 2k \), the series can be grouped as:
\[ S_n = (1 - 3) + (5 - 7) + (9 - 11) + \ldots + (4k - 3 - (4k - 1)) \]
- If \( n \) is odd, say \( n = 2k + 1 \), the last term will be positive:
\[ S_n = (1 - 3) + (5 - 7) + (9 - 11) + \ldots + (4k - 3) + (4k + 1) \]

### Step 3: Calculate the sum for even \( n \)
For \( n = 2k \):
\[ S_n = -2 + -2 + -2 + \ldots + (k \text{ times}) = -2k = -n \]

### Step 4: Calculate the sum for odd \( n \)
For \( n = 2k + 1 \):
\[ S_n = -2 + -2 + -2 + \ldots + (k \text{ times}) + (4k + 1) \]
This gives:
\[ S_n = -2k + (4k + 1) = 2k + 1 = n \]

### Step 5: Combine the results
Thus, we can summarize the result:
- If \( n \) is even, \( S_n = -\frac{n}{2} \)
- If \( n \) is odd, \( S_n = \frac{n + 1}{2} \)

### Final Result
The sum of the series \( S_n \) can be expressed as:
\[ S_n = \begin{cases} -\frac{n}{2} & \text{if } n \text{ is even} \\ \frac{n + 1}{2} & \text{if } n \text{ is odd} \end{cases} \]