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 \((n + 2r)^2\)
We start by expanding the square:
\[ (n + 2r)^2 = n^2 + 4nr + 4r^2 \]
Thus, we can rewrite the series as:
\[ S = \sum_{r=0}^{n} (-1)^r (n^2 + 4nr + 4r^2) \]

### Step 2: Separate the series
We can separate the summation into three parts:
\[ S = \sum_{r=0}^{n} (-1)^r n^2 + \sum_{r=0}^{n} (-1)^r 4nr + \sum_{r=0}^{n} (-1)^r 4r^2 \]
This simplifies to:
\[ S = n^2 \sum_{r=0}^{n} (-1)^r + 4n \sum_{r=0}^{n} (-1)^r r + 4 \sum_{r=0}^{n} (-1)^r r^2 \]

### Step 3: Evaluate the first summation
The first summation \( \sum_{r=0}^{n} (-1)^r \) is:
- If \( n \) is even, it equals \( 1 \) (since there are equal numbers of \( 1 \) and \( -1 \)).
- If \( n \) is odd, it equals \( 0 \).

Since \( n \) is even, we have:
\[ \sum_{r=0}^{n} (-1)^r = 1 \]
Thus:
\[ n^2 \sum_{r=0}^{n} (-1)^r = n^2 \cdot 1 = n^2 \]

### Step 4: Evaluate the second summation
For the second summation \( \sum_{r=0}^{n} (-1)^r r \):
- This can be calculated using the formula for the sum of an alternating series. The result is \( 0 \) when \( n \) is even.

Thus:
\[ 4n \sum_{r=0}^{n} (-1)^r r = 4n \cdot 0 = 0 \]

### Step 5: Evaluate the third summation
For the third summation \( \sum_{r=0}^{n} (-1)^r r^2 \):
- The formula for the sum of squares in an alternating series gives us \( \frac{n(n+1)}{2} \) when \( n \) is even.

Thus:
\[ 4 \sum_{r=0}^{n} (-1)^r r^2 = 4 \cdot \frac{n(n+1)}{2} = 2n(n+1) \]

### Step 6: Combine all parts
Now, we combine all the parts:
\[ S = n^2 + 0 + 2n(n+1) \]
This simplifies to:
\[ S = n^2 + 2n^2 + 2n = 3n^2 + 2n \]

### Final Answer
Thus, the sum of the series is:
\[ \boxed{3n^2 + 2n} \]