Question

If \( f(x) = \frac{2^x}{2^x + \sqrt{2}} \), \(x \in \mathbb{R}\), then \(\sum_{k=1}^{81} f\left(\frac{k}{82}\right)\) is equal to:

• A. 82
• B. \(\frac{81}{2}\)
• C. 41
• D. \(81\sqrt{2}\)

Hint:

Consider properties of exponential functions and symmetry when simplifying summations or integrals.

Answer 1

The Correct Option is B

To solve the problem, we need to find the value of the sum \( \sum_{k=1}^{81} f\left(\frac{k}{82}\right) \) where \( f(x) = \frac{2^x}{2^x + \sqrt{2}} \).

1. Analyze \( f(x) + f(1-x) \):
We compute \( f(1-x) \):
\( f(1-x) = \frac{2^{1-x}}{2^{1-x} + \sqrt{2}} = \frac{2/2^x}{2/2^x + \sqrt{2}} = \frac{2}{2 + \sqrt{2} \cdot 2^x} = \frac{\sqrt{2}}{\sqrt{2} + 2^x} \) 
Then we add \( f(x) \) and \( f(1-x) \):
\( f(x) + f(1-x) = \frac{2^x}{2^x + \sqrt{2}} + \frac{\sqrt{2}}{\sqrt{2} + 2^x} = \frac{2^x + \sqrt{2}}{2^x + \sqrt{2}} = 1 \)

2. Pair terms in the sum:
We pair the terms in the sum such that their arguments add up to 1. In other words, we pair \( f\left(\frac{k}{82}\right) \) with \( f\left(\frac{82-k}{82}\right) \):
\( f\left(\frac{k}{82}\right) + f\left(\frac{82-k}{82}\right) = f\left(\frac{k}{82}\right) + f\left(1 - \frac{k}{82}\right) = 1 \)

3. Rewrite the sum:
The original sum is \( \sum_{k=1}^{81} f\left(\frac{k}{82}\right) \). We can rewrite the sum as:
\( S = \sum_{k=1}^{40} \left[ f\left(\frac{k}{82}\right) + f\left(\frac{82-k}{82}\right) \right] + f\left(\frac{41}{82}\right) \) 
Since \( \frac{82-k}{82} = 1 - \frac{k}{82} \), each paired term in the sum equals 1.

4. Evaluate the sum:
The sum becomes:
\( S = \sum_{k=1}^{40} 1 + f\left(\frac{41}{82}\right) = 40 + f\left(\frac{1}{2}\right) \) 
Now, we evaluate \( f\left(\frac{1}{2}\right) \):
\( f\left(\frac{1}{2}\right) = \frac{2^{1/2}}{2^{1/2} + \sqrt{2}} = \frac{\sqrt{2}}{\sqrt{2} + \sqrt{2}} = \frac{\sqrt{2}}{2\sqrt{2}} = \frac{1}{2} \) 
So, \( S = 40 + \frac{1}{2} = \frac{81}{2} = 40.5 \)

Final Answer:
The value of the sum is \( \frac{81}{2} \).

Answer 2

Step 1: Write the given function.
We are given:
\[ f(x) = \frac{2^x}{2^x + \sqrt{2}}, \quad x \in \mathbb{R}. \]

Step 2: Find the complementary property.
Let’s calculate \( f(1 - x) \):
\[ f(1 - x) = \frac{2^{1 - x}}{2^{1 - x} + \sqrt{2}} = \frac{\frac{2}{2^x}}{\frac{2}{2^x} + \sqrt{2}}. \] Multiply numerator and denominator by \( 2^x \):
\[ f(1 - x) = \frac{2}{2 + \sqrt{2} \cdot 2^x}. \]

Step 3: Simplify and find the relationship between \( f(x) \) and \( f(1 - x) \).
To find a useful relationship, we add \( f(x) + f(1 - x) \):
\[ f(x) + f(1 - x) = \frac{2^x}{2^x + \sqrt{2}} + \frac{2}{2 + \sqrt{2} \cdot 2^x}. \] Multiply the second fraction numerator and denominator by \( 2^x \) for simplification:
\[ f(1 - x) = \frac{2 \cdot 2^x}{2 \cdot 2^x + \sqrt{2} \cdot 2^{2x}} = \frac{2 \cdot 2^x}{2^x(2 + \sqrt{2} \cdot 2^x)} = \frac{2}{2 + \sqrt{2} \cdot 2^x}. \] Now add them with a common denominator or observe numerically — the sum simplifies to 1:
\[ f(x) + f(1 - x) = 1. \]

Step 4: Apply this property to the summation.
We need: \[ S = \sum_{k=1}^{81} f\left(\frac{k}{82}\right). \] Notice that for each term \( f\left(\frac{k}{82}\right) \), there exists another term \( f\left(1 - \frac{k}{82}\right) = f\left(\frac{82 - k}{82}\right) \).
For \( k = 1, 2, 3, \dots, 81 \), pairs \( \left(\frac{k}{82}, \frac{82 - k}{82}\right) \) exist such that:
\[ f\left(\frac{k}{82}\right) + f\left(\frac{82 - k}{82}\right) = 1. \]

Step 5: Count the number of pairs.
There are 81 terms, and since they form 40 such pairs (since 81 is odd, one term corresponds to \( k = 41 \) being its own complement). Thus:
\[ S = 40 \times 1 + f\left(\frac{41}{82}\right). \] But \( f\left(\frac{41}{82}\right) + f\left(1 - \frac{41}{82}\right) = 1 \), and both terms are the same since they are equal when \( x = \frac{1}{2} \). Hence \( f\left(\frac{41}{82}\right) = \frac{1}{2} \).

Step 6: Final sum.
\[ S = 40 + \frac{1}{2} = \frac{81}{2}. \]

Final Answer:
\[ \boxed{\frac{81}{2}} \]