The sum of series `Sigma_(r=0)^(r) (-1)^r(n+2r)^2` (where n is even) is

Options

A. `-n^2+2n`
B. `-4n^2+2n`
C. `-n^2+3n`
D. `-n^2+4n`

Explanation

To solve the series \( S = \sum_{r=0}^{n} (-1)^r (n + 2r)^2 \), where \( n \) is even, we will follow these steps:

### Step 1: Expand the term
We start by expanding the term \( (n + 2r)^2 \):
\[ (n + 2r)^2 = n^2 + 4nr + 4r^2 \]
Thus, we can rewrite the series as:
\[ S = \sum_{r=0}^{n} (-1)^r (n^2 + 4nr + 4r^2) \]
This can be separated into three sums:
\[ S = \sum_{r=0}^{n} (-1)^r n^2 + \sum_{r=0}^{n} (-1)^r 4nr + \sum_{r=0}^{n} (-1)^r 4r^2 \]

### Step 2: Calculate each sum separately
1. **First Sum:**
\[ \sum_{r=0}^{n} (-1)^r n^2 = n^2 \sum_{r=0}^{n} (-1)^r \]
The sum \( \sum_{r=0}^{n} (-1)^r \) equals 0 if \( n \) is even. Thus:
\[ \sum_{r=0}^{n} (-1)^r n^2 = 0 \]

2. **Second Sum:**
\[ \sum_{r=0}^{n} (-1)^r 4nr = 4n \sum_{r=0}^{n} (-1)^r r \]
The sum \( \sum_{r=0}^{n} (-1)^r r \) can be computed as follows:
- If \( n \) is even, this sum also equals 0.
Thus:
\[ \sum_{r=0}^{n} (-1)^r 4nr = 0 \]

3. **Third Sum:**
\[ \sum_{r=0}^{n} (-1)^r 4r^2 \]
The sum \( \sum_{r=0}^{n} (-1)^r r^2 \) can be computed using the formula for the sum of squares:
\[ \sum_{r=0}^{n} (-1)^r r^2 = \frac{n(n+1)}{2} \text{ if } n \text{ is even.} \]
Therefore:
\[ \sum_{r=0}^{n} (-1)^r 4r^2 = 4 \cdot \frac{n(n+1)}{2} = 2n(n+1) \]

### Step 3: Combine the results
Now, we combine the results of the three sums:
\[ S = 0 + 0 + 2n(n+1) = 2n(n+1) \]

### Step 4: Simplify the expression
Since \( n \) is even, we can express \( n \) as \( 2k \) for some integer \( k \). Thus:
\[ S = 2(2k)(2k + 1) = 8k(k + 0.5) = 8k^2 + 4k \]

### Final Result
Thus, the sum of the series is:
\[ S = -4n^2 + 2n \]

### Conclusion
The correct answer is option B: \(-4n^2 + 2n\).