Solveeit Logo
Login

Question

Question: If $f(x)=\frac{3^{1-x}}{3^{1-x}+3^x}$ for all $x \in Q$, then the value of the sum $f(\frac{1}{55})+...

If f(x)=31−x31−x+3xf(x)=\frac{3^{1-x}}{3^{1-x}+3^x} for all x∈Qx \in Q, then the value of the sum f(155)+f(255)+...+f(5455)f(\frac{1}{55})+f(\frac{2}{55})+...+f(\frac{54}{55}) is ____.

Answer

27

Explanation

Solution

The problem asks for the sum of a series involving a function f(x)f(x). The key to solving such problems is often to find a useful property of the function, typically f(x)+f(a−x)f(x) + f(a-x) or f(x)+f(1/x)f(x) + f(1/x).

Given the function f(x)=31−x31−x+3xf(x)=\frac{3^{1-x}}{3^{1-x}+3^x}.

Step 1: Find the property of the function f(x)f(x).

Let's evaluate f(1−x)f(1-x):

f(1−x)=31−(1−x)31−(1−x)+31−xf(1-x) = \frac{3^{1-(1-x)}}{3^{1-(1-x)}+3^{1-x}}

f(1−x)=31−1+x31−1+x+31−xf(1-x) = \frac{3^{1-1+x}}{3^{1-1+x}+3^{1-x}}

f(1−x)=3x3x+31−xf(1-x) = \frac{3^x}{3^x+3^{1-x}}

Now, let's add f(x)f(x) and f(1−x)f(1-x):

f(x)+f(1−x)=31−x31−x+3x+3x3x+31−xf(x) + f(1-x) = \frac{3^{1-x}}{3^{1-x}+3^x} + \frac{3^x}{3^x+3^{1-x}}

Since the denominators are the same, we can add the numerators:

f(x)+f(1−x)=31−x+3x31−x+3xf(x) + f(1-x) = \frac{3^{1-x} + 3^x}{3^{1-x}+3^x}

f(x)+f(1−x)=1f(x) + f(1-x) = 1

This property is crucial for evaluating the sum.

Step 2: Evaluate the sum.

The sum to be evaluated is S=f(155)+f(255)+...+f(5455)S = f(\frac{1}{55})+f(\frac{2}{55})+...+f(\frac{54}{55}).

There are 5454 terms in this sum.

We can pair the terms such that each pair sums to 1 using the property f(x)+f(1−x)=1f(x) + f(1-x) = 1.

Consider a general term f(k55)f\left(\frac{k}{55}\right). Its pair will be f(1−k55)=f(55−k55)f\left(1-\frac{k}{55}\right) = f\left(\frac{55-k}{55}\right).

So, f(k55)+f(55−k55)=1f\left(\frac{k}{55}\right) + f\left(\frac{55-k}{55}\right) = 1.

Let's list the pairs:

The first term f(155)f(\frac{1}{55}) is paired with the last term f(5455)f(\frac{54}{55}):

f(155)+f(5455)=f(155)+f(1−155)=1f(\frac{1}{55}) + f(\frac{54}{55}) = f(\frac{1}{55}) + f(1-\frac{1}{55}) = 1.

The second term f(255)f(\frac{2}{55}) is paired with the second to last term f(5355)f(\frac{53}{55}):

f(255)+f(5355)=f(255)+f(1−255)=1f(\frac{2}{55}) + f(\frac{53}{55}) = f(\frac{2}{55}) + f(1-\frac{2}{55}) = 1.

This pairing continues. Since there are 5454 terms in total, and 5454 is an even number, all terms can be perfectly paired up.

The number of such pairs is Total number of terms2=542=27\frac{\text{Total number of terms}}{2} = \frac{54}{2} = 27.

Each pair sums to 1.

Therefore, the total sum SS is the number of pairs multiplied by the sum of each pair:

S=27×1=27S = 27 \times 1 = 27

The value of the sum is 27.