Question

Let $f:[0,1]\rightarrow R$ be defined by $f(x)=\frac{4^x}{4^x+2}$ then value of $2\left[f(\frac{1}{40})+f(\frac{2}{40})+f(\frac{3}{40})+...+f(\frac{39}{40})\right]$

Answer 1

39

Answer 2

The function is given by $f(x) = \frac{4^x}{4^x+2}$ for $x \in [0,1]$.
We first examine the property of the function $f(x)$ with respect to $1-x$.
$f(1-x) = \frac{4^{1-x}}{4^{1-x}+2} = \frac{\frac{4}{4^x}}{\frac{4}{4^x}+2} = \frac{\frac{4}{4^x}}{\frac{4+2 \cdot 4^x}{4^x}} = \frac{4}{4+2 \cdot 4^x} = \frac{2}{2+4^x}$.
Now, let's add $f(x)$ and $f(1-x)$:
$f(x) + f(1-x) = \frac{4^x}{4^x+2} + \frac{2}{4^x+2} = \frac{4^x+2}{4^x+2} = 1$.
So, $f(x) + f(1-x) = 1$ for all $x \in [0,1]$.

We need to evaluate the sum $S = f(\frac{1}{40})+f(\frac{2}{40})+f(\frac{3}{40})+...+f(\frac{39}{40})$.
The sum consists of 39 terms of the form $f(\frac{k}{40})$ for $k=1, 2, ..., 39$.
We can group the terms in pairs $(f(\frac{k}{40}), f(\frac{40-k}{40}))$.
For $k=1, 2, ..., 19$, the term $f(\frac{k}{40})$ is paired with $f(\frac{40-k}{40}) = f(1-\frac{k}{40})$.
Using the property $f(x) + f(1-x) = 1$, we have $f(\frac{k}{40}) + f(\frac{40-k}{40}) = 1$.
There are 19 such pairs corresponding to $k=1, 2, ..., 19$.
The sum of these 19 pairs is $19 \times 1 = 19$.

The sum $S$ can be written as:
$S = [f(\frac{1}{40})+f(\frac{39}{40})] + [f(\frac{2}{40})+f(\frac{38}{40})] + ... + [f(\frac{19}{40})+f(\frac{21}{40})] + f(\frac{20}{40})$.
The terms from $k=1$ to $k=19$ are paired with terms from $k=39$ down to $k=21$.
The middle term is when $k = 40-k$, which gives $2k=40$, so $k=20$. The middle term is $f(\frac{20}{40}) = f(\frac{1}{2})$. This term is not part of any pair summing to 1 within the sum.

So, the sum is $S = \sum_{k=1}^{19} \left[f(\frac{k}{40}) + f(\frac{40-k}{40})\right] + f(\frac{20}{40})$.
$S = \sum_{k=1}^{19} 1 + f(\frac{1}{2})$.
$S = 19 \times 1 + f(\frac{1}{2})$.
$S = 19 + f(\frac{1}{2})$.

Now we calculate the value of $f(\frac{1}{2})$:
$f(\frac{1}{2}) = \frac{4^{1/2}}{4^{1/2}+2} = \frac{\sqrt{4}}{\sqrt{4}+2} = \frac{2}{2+2} = \frac{2}{4} = \frac{1}{2}$.

Substitute the value of $f(\frac{1}{2})$ back into the expression for $S$:
$S = 19 + \frac{1}{2} = \frac{38+1}{2} = \frac{39}{2}$.

The question asks for the value of $2 \left[f(\frac{1}{40})+f(\frac{2}{40})+f(\frac{3}{40})+...+f(\frac{39}{40})\right]$, which is $2S$.
$2S = 2 \times \frac{39}{2} = 39$.

The final answer is $\boxed{39}$.