If `f(x)=(4^(x))/(4^(x)+2)," then "f(1/(97))+f((2)/(97))+...+f((96)/(97))` is equal to:

Options

A. 1
B. 48
C. `-48`
D. `-1`

Answer

To solve the problem, we need to evaluate the sum \( f\left(\frac{1}{97}\right) + f\left(\frac{2}{97}\right) + \ldots + f\left(\frac{96}{97}\right) \) where \( f(x) = \frac{4^x}{4^x + 2} \).

### Step-by-Step Solution:

1. **Define the Function**:
\[ f(x) = \frac{4^x}{4^x + 2} \]

2. **Find \( f(1 - x) \)**:
We need to find \( f(1 - x) \):
\[ f(1 - x) = \frac{4^{1 - x}}{4^{1 - x} + 2} = \frac{4 \cdot 4^{-x}}{4 \cdot 4^{-x} + 2} = \frac{4}{4 + 2 \cdot 4^x} \]

3. **Simplify \( f(1 - x) \)**:
We can rewrite \( f(1 - x) \) as:
\[ f(1 - x) = \frac{4}{2(2 + 4^x)} = \frac{2}{2 + 4^x} \]

4. **Add \( f(x) \) and \( f(1 - x) \)**:
Now, we add \( f(x) \) and \( f(1 - x) \):
\[ f(x) + f(1 - x) = \frac{4^x}{4^x + 2} + \frac{2}{2 + 4^x} \]
The denominators are the same, so we can combine them:
\[ f(x) + f(1 - x) = \frac{4^x + 2}{4^x + 2} = 1 \]

5. **Pairing Terms**:
Notice that \( f\left(\frac{k}{97}\right) + f\left(\frac{97-k}{97}\right) = 1 \) for \( k = 1, 2, \ldots, 96 \). This means:
\[ f\left(\frac{1}{97}\right) + f\left(\frac{96}{97}\right) = 1 \]
\[ f\left(\frac{2}{97}\right) + f\left(\frac{95}{97}\right) = 1 \]
Continuing this way, we can pair the terms up to \( k = 48 \).

6. **Count the Pairs**:
Since there are 96 terms in total, we can form \( \frac{96}{2} = 48 \) pairs, each summing to 1.

7. **Calculate the Total**:
Therefore, the total sum is:
\[ f\left(\frac{1}{97}\right) + f\left(\frac{2}{97}\right) + \ldots + f\left(\frac{96}{97}\right) = 48 \cdot 1 = 48 \]

### Final Answer:
\[ \boxed{48} \]