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: Function Analysis.
Observe that \( f(x) = \frac{2^x}{2^x + \sqrt{2}} \) simplifies. For large \(x\), the exponential term dominates, causing \(f(x)\) to approach 1.
Step 2: Value Calculation.
Calculate \( f\left(\frac{k}{82}\right) \) for \( k = 1 \) to \( 81 \). Note the function's symmetry about \( x = 0 \).
Step 3: Symmetry Application.
The symmetry of \( f(x) \) about \( x = 0 \) simplifies the summation. Sum terms using the midpoint Riemann sum approximation for the integral of the function over \( [0, 1] \).
Step 4: Sum Computation.
Sum the values. \( \sum_{k=1}^{81} f\left(\frac{k}{82}\right) \) approaches \( \frac{81}{2} \) as \( k \) reaches 81.
Conclusion: The sum evaluates to \( \frac{81}{2} \), illustrating the impact of exponential function growth on the sum.