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 expression \( f\left(\frac{1}{97}\right) + f\left(\frac{2}{97}\right) + \ldots + f\left(\frac{96}{97}\right) \) where \( f(x) = \frac{4^x}{4^x + 2} \).

### Step 1: Find \( f(-x) \)

We start by calculating \( f(-x) \):

\[ f(-x) = \frac{4^{-x}}{4^{-x} + 2} \]

This can be rewritten as:

\[ f(-x) = \frac{\frac{1}{4^x}}{\frac{1}{4^x} + 2} = \frac{1}{1 + 2 \cdot 4^x} \]

### Step 2: Find \( f(x) + f(-x) \)

Next, we will find the sum \( f(x) + f(-x) \):

\[ f(x) + f(-x) = \frac{4^x}{4^x + 2} + \frac{1}{1 + 2 \cdot 4^x} \]

To combine these fractions, we need a common denominator:

\[ = \frac{4^x(1 + 2 \cdot 4^x) + (4^x + 2)}{(4^x + 2)(1 + 2 \cdot 4^x)} \]

Expanding the numerator:

\[ = \frac{4^x + 2 \cdot 4^{2x} + 4^x + 2}{(4^x + 2)(1 + 2 \cdot 4^x)} = \frac{2 \cdot 4^x + 2 \cdot 4^{2x} + 2}{(4^x + 2)(1 + 2 \cdot 4^x)} \]

Factoring out 2 from the numerator:

\[ = \frac{2(4^{2x} + 4^x + 1)}{(4^x + 2)(1 + 2 \cdot 4^x)} \]

### Step 3: Simplifying \( f(x) + f(-x) \)

Notice that \( 4^{2x} + 4^x + 1 \) can be factored as \( (4^x + 1)^2 \):

\[ = \frac{2(4^x + 1)^2}{(4^x + 2)(1 + 2 \cdot 4^x)} \]

This shows that \( f(x) + f(-x) = 1 \).

### Step 4: Evaluate the sum

Now, we can use the property we found:

\[ f\left(\frac{k}{97}\right) + f\left(\frac{97-k}{97}\right) = 1 \]

for \( k = 1, 2, \ldots, 48 \).

The pairs are:

- \( f\left(\frac{1}{97}\right) + f\left(\frac{96}{97}\right) = 1 \)
- \( f\left(\frac{2}{97}\right) + f\left(\frac{95}{97}\right) = 1 \)
- ...
- \( f\left(\frac{48}{97}\right) + f\left(\frac{49}{97}\right) = 1 \)

### Step 5: Count the pairs

There are 48 pairs in total, and each pair sums to 1:

\[ \text{Total sum} = 48 \cdot 1 = 48 \]

### Final Answer

Thus, the value of \( f\left(\frac{1}{97}\right) + f\left(\frac{2}{97}\right) + \ldots + f\left(\frac{96}{97}\right) \) is:

\[ \boxed{48} \]