Let the functions `f: [0,1] to R ` be defined by
`f(x) = (4^(x))/(4^(x)+2)` Then the value of
`f(1/40)+f(2/20) +f(3/40) +....+ (39/40) - f(1/2)` is _______

Answer

To solve the problem, we need to evaluate the expression:

\[ f\left(\frac{1}{40}\right) + f\left(\frac{2}{40}\right) + f\left(\frac{3}{40}\right) + \ldots + f\left(\frac{39}{40}\right) - f\left(\frac{1}{2}\right) \]

where the function \( f(x) \) is defined as:

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

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

First, we calculate \( f(1 - x) \):

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

This can be rewritten as:

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

### Step 2: Simplify \( f(x) + f(1 - x) \)

Now, we will add \( f(x) \) and \( f(1 - x) \):

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

Substituting \( f(1 - x) \):

\[ = \frac{4^x}{4^x + 2} + \frac{4 \cdot 4^{-x}}{4 \cdot 4^{-x} + 2} \]

### Step 3: Find a common denominator

The common denominator is \( (4^x + 2)(4 \cdot 4^{-x} + 2) \):

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

### Step 4: Simplify the numerator

Expanding the numerator:

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

### Step 5: Show that \( f(x) + f(1 - x) = 1 \)

We can see that the numerator simplifies to \( 4 + 2 \cdot (4^x + 4^{-x}) \) and the denominator simplifies to \( 4^x + 2 + 2 + 4^{-x} = 4 + 2 + 4^{-x} + 2 = 4 + 2 = 6 \).

Thus, we conclude that:

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

### Step 6: Pairing terms

Now, we can pair the terms in the original sum:

\[ f\left(\frac{k}{40}\right) + f\left(\frac{40 - k}{40}\right) = 1 \]

for \( k = 1, 2, \ldots, 19 \). Each pair sums to 1, and there are 19 pairs:

\[ f\left(\frac{1}{40}\right) + f\left(\frac{39}{40}\right) = 1 \]
\[ f\left(\frac{2}{40}\right) + f\left(\frac{38}{40}\right) = 1 \]
\[ \ldots \]
\[ f\left(\frac{19}{40}\right) + f\left(\frac{21}{40}\right) = 1 \]

### Step 7: Calculate the total

Thus, we have 19 pairs contributing 19 to the sum:

\[ f\left(\frac{1}{40}\right) + f\left(\frac{2}{40}\right) + \ldots + f\left(\frac{39}{40}\right) = 19 \]

Now we subtract \( f\left(\frac{1}{2}\right) \):

\[ f\left(\frac{1}{2}\right) = \frac{4^{1/2}}{4^{1/2} + 2} = \frac{2}{2 + 2} = \frac{2}{4} = \frac{1}{2} \]

### Final Calculation

Thus, the final value is:

\[ 19 - \frac{1}{2} = 19 - 0.5 = 18.5 \]

### Final Answer

The value of the expression is:

\[ \boxed{18.5} \]