Let f(x)=2007^x/(2007^x+2007^1/2), then the sum f(1/2007)+f(2/2007)+f(3/2007)+....+f(2006/2007) is equal to

(1) 1004 (2) 1003

(3) 1003.5 (4) 1004.5

To solve the problem, we first need to analyze the function given: \( f(x) = \frac{2007^x}{2007^x + 2007^{1/2}} \). We can simplify this function to make it easier to work with. Let's rewrite it in a more manageable form.

Function Simplification

We can factor out \( 2007^{1/2} \) from the denominator:

\( f(x) = \frac{2007^x}{2007^x + 2007^{1/2}} = \frac{2007^x}{2007^{1/2}(2007^{x - 1/2} + 1)} = \frac{2007^{x - 1/2}}{2007^{x - 1/2} + 1} \)

This shows that \( f(x) \) can be expressed as a function of \( 2007^{x - 1/2} \).

Finding the Complementary Function

Next, let's find \( f(1 - x) \) to see if there's a relationship between \( f(x) \) and \( f(1 - x) \):

\( f(1 - x) = \frac{2007^{1 - x}}{2007^{1 - x} + 2007^{1/2}} = \frac{2007^{1 - x}}{2007^{1/2}(2007^{1 - x - 1/2} + 1)} = \frac{2007^{1/2} \cdot 2007^{-x}}{2007^{1/2} \cdot 2007^{-x} + 1} = \frac{2007^{1/2}}{2007^{1/2} + 2007^{x}} \)

Now, we can see that:

\( f(x) + f(1 - x) = 1 \)

This symmetry will be crucial for our calculations.

Summing the Function Values

We need to calculate the sum:

\( S = f\left(\frac{1}{2007}\right) + f\left(\frac{2}{2007}\right) + f\left(\frac{3}{2007}\right) + \ldots + f\left(\frac{2006}{2007}\right) \)

Notice that for each \( k \) from 1 to 2006, we can pair \( f\left(\frac{k}{2007}\right) \) with \( f\left(\frac{2007 - k}{2007}\right) \). Each pair sums to 1:

• For \( k = 1 \), \( f\left(\frac{1}{2007}\right) + f\left(\frac{2006}{2007}\right) = 1 \)
• For \( k = 2 \), \( f\left(\frac{2}{2007}\right) + f\left(\frac{2005}{2007}\right) = 1 \)
• And so on, until \( k = 1003 \).

Counting the Pairs

Since we have 2006 terms, we can form 1003 pairs, each summing to 1:

\( S = 1003 \times 1 = 1003 \)

Final Result

Thus, the sum \( f\left(\frac{1}{2007}\right) + f\left(\frac{2}{2007}\right) + \ldots + f\left(\frac{2006}{2007}\right) \) equals 1003. Therefore, the correct answer is:

(2) 1003