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 \( \frac{81}{2} \).