Let `f(k)=k/2009` and `g(k)=(f^4(k))/((1-f(k))^4+(f(k))^4)` then the sum `sum_(k=0)^2009g(k)` is equal:

Options

A. 2009
B. 2008
C. 1005
D. 1004

Answer

To solve the problem step-by-step, we start with the definitions given in the question:

1. **Define the functions**:
- Let \( f(k) = \frac{k}{2009} \)
- Let \( g(k) = \frac{f(k)^4}{(1 - f(k))^4 + f(k)^4} \)

2. **Substituting \( f(k) \) into \( g(k) \)**:
- Substitute \( f(k) \) into \( g(k) \):
\[ g(k) = \frac{\left(\frac{k}{2009}\right)^4}{\left(1 - \frac{k}{2009}\right)^4 + \left(\frac{k}{2009}\right)^4} \]

3. **Simplifying \( g(k) \)**:
- We can express \( g(k) \) as:
\[ g(k) = \frac{\frac{k^4}{2009^4}}{\left(\frac{2009 - k}{2009}\right)^4 + \frac{k^4}{2009^4}} = \frac{k^4}{(2009 - k)^4 + k^4} \cdot \frac{1}{2009^4} \]

4. **Finding the sum \( \sum_{k=0}^{2009} g(k) \)**:
- We need to compute:
\[ \sum_{k=0}^{2009} g(k) = \sum_{k=0}^{2009} \frac{k^4}{(2009 - k)^4 + k^4} \cdot \frac{1}{2009^4} \]

5. **Using symmetry in the sum**:
- Notice that \( g(2009 - k) \) can be expressed as:
\[ g(2009 - k) = \frac{(2009 - k)^4}{k^4 + (2009 - k)^4} \]
- Therefore, we can pair \( g(k) \) and \( g(2009 - k) \):
\[ g(k) + g(2009 - k) = 1 \]

6. **Counting the pairs**:
- The sum \( \sum_{k=0}^{2009} g(k) \) can be grouped into pairs:
\[ (g(0) + g(2009)) + (g(1) + g(2008)) + \ldots + (g(1004) + g(1005)) \]
- There are 1005 pairs, and each pair sums to 1.

7. **Calculating the total sum**:
- Thus, the total sum is:
\[ \sum_{k=0}^{2009} g(k) = 1005 \]

8. **Final answer**:
- Therefore, the sum \( \sum_{k=0}^{2009} g(k) \) is equal to \( 1005 \).