Question

Consider $f(x) = \frac{4^x}{4^x+2}$, if $f(\frac{1}{1997})+f(\frac{2}{1997})+ \dots + f(\frac{1996}{1997}) = 499q$, then q is equal to

Answer 1

2

Answer 2

To solve the problem, we first analyze the given function $f(x) = \frac{4^x}{4^x+2}$.

We look for a special property of this function, specifically $f(x) + f(1-x)$.

1. Find $f(1-x)$:

   Substitute $(1-x)$ for $x$ in the function definition: $f(1-x) = \frac{4^{1-x}}{4^{1-x}+2}$

   We can rewrite $4^{1-x}$ as $4 \cdot 4^{-x} = \frac{4}{4^x}$.

   So, $f(1-x) = \frac{\frac{4}{4^x}}{\frac{4}{4^x}+2}$

   To simplify, multiply the numerator and denominator by $4^x$:

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

   Factor out 2 from the denominator:

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

2. Calculate $f(x) + f(1-x)$:

   Now, add $f(x)$ and $f(1-x)$:

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

   Since both terms have the same denominator, we can add the numerators:

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

   This property, $f(x) + f(1-x) = 1$, is crucial for solving the problem.

3. Evaluate the sum:

   The given sum is $S = f(\frac{1}{1997})+f(\frac{2}{1997})+ \dots + f(\frac{1996}{1997})$.

   Let $N = 1997$. The sum can be written as $S = \sum_{k=1}^{N-1} f\left(\frac{k}{N}\right)$.

   The terms in the sum range from $k=1$ to $k=1996$. There are $1996$ terms in total.

   We can pair the terms using the property $f(x) + f(1-x) = 1$.

   Consider a pair of terms $f\left(\frac{k}{N}\right)$ and $f\left(\frac{N-k}{N}\right)$.

   Note that $\frac{N-k}{N} = 1 - \frac{k}{N}$.

   So, $f\left(\frac{k}{N}\right) + f\left(\frac{N-k}{N}\right) = f\left(\frac{k}{N}\right) + f\left(1-\frac{k}{N}\right) = 1$.

   The terms in the sum can be paired as follows:

   $(f(\frac{1}{1997}) + f(\frac{1996}{1997})) + (f(\frac{2}{1997}) + f(\frac{1995}{1997})) + \dots$

   Since there are $1996$ terms, and $1996$ is an even number, all terms can be perfectly paired.

   The number of such pairs is $\frac{1996}{2} = 998$.

   Each pair sums to 1.

   Therefore, the total sum $S = 998 \times 1 = 998$.

4. Solve for q:

   We are given that the sum $S = 499q$.

   We found $S = 998$.

   So, $998 = 499q$.

   To find $q$, divide both sides by $499$:

   $q = \frac{998}{499}$

   $q = 2$

The final answer is $\boxed{2}$.

Explanation of the solution:

The function $f(x) = \frac{4^x}{4^x+2}$ possesses the property $f(x) + f(1-x) = 1$. This is derived by simplifying $f(1-x)$ to $\frac{2}{4^x+2}$ and adding it to $f(x)$. The given sum consists of $1996$ terms of the form $f(\frac{k}{1997})$. By pairing terms $f(\frac{k}{1997})$ with $f(\frac{1997-k}{1997})$, each pair sums to $f(\frac{k}{1997}) + f(1-\frac{k}{1997}) = 1$. Since there are $1996$ terms, there are $1996/2 = 998$ such pairs. Thus, the total sum is $998 \times 1 = 998$. Equating this to $499q$, we get $998 = 499q$, which yields $q=2$.