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.  
Notice that \( f(x) = \frac{2^x}{2^x + \sqrt{2}} \) simplifies because the exponential term becomes dominant for large \(x\) values, and the function approaches 1 as \(x\) increases.
Step 2: Compute individual values. 
Evaluate \( f\left(\frac{k}{82}\right) \) for each \( k \) from 1 to 81, paying attention to the symmetry of the function around \( x = 0 \).
Step 3: Apply symmetry in calculation. 
Thanks to the symmetry of \( f(x) \) around \( x = 0 \), simplifications can be made in the summation process. Use the midpoint Riemann sum approximation for the integral of the function over the interval \( [0, 1] \).
Step 4: Calculate the total sum. 
After summing the individual terms, the result \( \sum_{k=1}^{81} f\left(\frac{k}{82}\right) \) approaches \( \frac{81}{2} \) as \( k \) increases to 81.
Conclusion: The final sum is \( \frac{81}{2} \), showing how the exponential function's behavior affects the total sum.