`f(x) = 9^x/ ( 9 + 9^x)` then value of `f(1/2015) + f(2/2015) .......+ f(4029/2015)`

Options

A. 1007
B. `(4029)/(2)`
C. 2014
D. 2015

Explanation

To solve the problem step by step, we start with the function given:

**Step 1:** Define the function.
\[ f(x) = \frac{9^x}{9 + 9^x} \]

**Step 2:** Simplify the function.
We can rewrite \( f(x) \) by dividing the numerator and denominator by \( 9^x \):
\[ f(x) = \frac{9^x}{9 + 9^x} = \frac{1}{\frac{9}{9^x} + 1} = \frac{1}{\frac{9}{9^x} + 1} = \frac{1}{\frac{9}{9^x} + 1} = \frac{1}{9^{1-x} + 1} \]

**Step 3:** Calculate \( f\left(\frac{k}{2015}\right) \) for \( k = 1, 2, \ldots, 4029 \).
We will find \( f\left(\frac{k}{2015}\right) \):
\[ f\left(\frac{k}{2015}\right) = \frac{1}{9^{1 - \frac{k}{2015}} + 1} = \frac{1}{9^{\frac{2015 - k}{2015}} + 1} \]

**Step 4:** Find \( f\left(\frac{4029 - k}{2015}\right) \).
Now, we can also calculate:
\[ f\left(\frac{4029 - k}{2015}\right) = \frac{1}{9^{1 - \frac{4029 - k}{2015}} + 1} = \frac{1}{9^{\frac{4029 - 2015 + k}{2015}} + 1} = \frac{1}{9^{\frac{2014 + k}{2015}} + 1} \]

**Step 5:** Add \( f\left(\frac{k}{2015}\right) \) and \( f\left(\frac{4029 - k}{2015}\right) \).
Now, we can observe that:
\[ f\left(\frac{k}{2015}\right) + f\left(\frac{4029 - k}{2015}\right) = \frac{1}{9^{\frac{2015 - k}{2015}} + 1} + \frac{1}{9^{\frac{2014 + k}{2015}} + 1} \]

**Step 6:** Simplify the sum.
Using the property of the function:
\[ f(x) + f(1-x) = 1 \]
We can conclude that:
\[ f\left(\frac{k}{2015}\right) + f\left(\frac{4029 - k}{2015}\right) = 1 \]

**Step 7:** Count the number of pairs.
Since \( k \) ranges from \( 1 \) to \( 4029 \), we have \( 4029 \) terms. The pairs will be \( (1, 4028), (2, 4027), \ldots, (2014, 2015) \), which gives us \( 2014 \) pairs plus the middle term \( f(2015/2015) = f(1) \).

**Step 8:** Calculate \( f(1) \).
\[ f(1) = \frac{9^1}{9 + 9^1} = \frac{9}{9 + 9} = \frac{9}{18} = \frac{1}{2} \]

**Step 9:** Calculate the total sum.
The total sum is:
\[ 2014 \cdot 1 + \frac{1}{2} = 2014 + \frac{1}{2} = 2014.5 = \frac{4029}{2} \]

Thus, the final answer is:
\[ \boxed{\frac{4029}{2}} \]