Find the sum of the series: `1^2-2^2+3^2-4^2+.....-2008^2+2009^2`.

Options

A. 2019045
B. 1005004
C. `2000506`
D. none of these

Answer

To find the sum of the series \( S = 1^2 - 2^2 + 3^2 - 4^2 + \ldots - 2008^2 + 2009^2 \), we can group the terms in pairs and simplify using the difference of squares formula.

### Step-by-Step Solution:

1. **Group the Terms**:
We can group the terms in pairs:
\[ S = (1^2 - 2^2) + (3^2 - 4^2) + (5^2 - 6^2) + \ldots + (2007^2 - 2008^2) + 2009^2 \]

2. **Apply the Difference of Squares**:
Recall the formula for the difference of squares:
\[ a^2 - b^2 = (a - b)(a + b) \]
For each pair, we have:
\[ n^2 - (n+1)^2 = n^2 - (n^2 + 2n + 1) = - (2n + 1) \]
Therefore, we can rewrite the series as:
\[ S = - (2 \cdot 1 + 1) - (2 \cdot 3 + 1) - (2 \cdot 5 + 1) - \ldots - (2 \cdot 2007 + 1) + 2009^2 \]

3. **Count the Number of Pairs**:
The pairs are formed from \( 1 \) to \( 2008 \), which gives us \( 1004 \) pairs (since there are \( 2008/2 = 1004 \) pairs).

4. **Sum the Pairs**:
The sum of each pair can be expressed as:
\[ S = -\sum_{k=0}^{1003} (2(2k + 1) + 1) + 2009^2 \]
This simplifies to:
\[ S = -\sum_{k=0}^{1003} (4k + 3) + 2009^2 \]

5. **Calculate the Sum**:
The sum can be split into two parts:
\[ S = -\left(4\sum_{k=0}^{1003} k + \sum_{k=0}^{1003} 3\right) + 2009^2 \]
Using the formula for the sum of the first \( n \) natural numbers:
\[ \sum_{k=0}^{n} k = \frac{n(n + 1)}{2} \]
For \( n = 1003 \):
\[ \sum_{k=0}^{1003} k = \frac{1003 \cdot 1004}{2} = 503006 \]
Therefore:
\[ S = -\left(4 \cdot 503006 + 3 \cdot 1004\right) + 2009^2 \]

6. **Calculate Each Component**:
\[ 4 \cdot 503006 = 2012024 \]
\[ 3 \cdot 1004 = 3012 \]
Thus:
\[ S = - (2012024 + 3012) + 2009^2 \]
\[ S = -2015036 + 2009^2 \]

7. **Calculate \( 2009^2 \)**:
\[ 2009^2 = 4036081 \]
Therefore:
\[ S = -2015036 + 4036081 = 2021045 \]

### Final Result:
The sum of the series is:
\[ \boxed{2021045} \]