If \( f(x) = \frac{2^x}{2^x + \sqrt{2}} , \ x \in \mathbb{R} \), then \( \displaystyle \sum_{k=1}^{81} f\!\left( \frac{k}{82} \right) \) is equal to:
\( 41 \)
\( 82 \)
\( 81\sqrt{2} \)
\( \frac{81}{2} \)