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

Step 1: Analyze the function.
Observe that \( f(x) = \frac{2^x}{2^x + \sqrt{2}} \) simplifies as the exponential function dominates for large \(x\) values, approaching 1.
Step 2: Calculate individual values.
Calculate \( f\left(\frac{k}{82}\right) \) for \( k = 1 \) to \( 81 \), noting the function's symmetry around \( x = 0 \).
Step 3: Utilize symmetry in calculation.
The symmetry of \( f(x) \) around \( x = 0 \) allows for simplifications in summation. Sum the terms using the midpoint Riemann sum approximation for the integral of the function over \( [0, 1] \).
Step 4: Compute the total sum.
Sum the values and find that \( \sum_{k=1}^{81} f\left(\frac{k}{82}\right) \) approaches \( \frac{81}{2} \) as \( k \) increases to 81.
Conclusion: The sum evaluates to \( \frac{81}{2} \), demonstrating how the exponential function's growth affects the sum.