If $$f(x)=\frac{2^{x}}{2^{x}+\sqrt{2}},x \in \mathbb{R}$$, then $$\sum_{k=1}^{81}f(\frac{k}{82})$$ is equals to

We are given $$f(x) = \frac{2^x}{2^x + \sqrt{2}}$$ and need to find $$\sum_{k=1}^{81} f\left(\frac{k}{82}\right)\,. $$

First observe that $$f(1-x) = \frac{2^{1-x}}{2^{1-x} + \sqrt{2}} = \frac{2\cdot 2^{-x}}{2\cdot 2^{-x} + \sqrt{2}}\,. $$ Multiplying numerator and denominator by $$2^x$$ gives $$f(1-x) = \frac{2}{2 + \sqrt{2}\cdot 2^x}\,. $$ Hence $$ f(x) + f(1-x) = \frac{2^x}{2^x + \sqrt{2}} + \frac{2}{2 + \sqrt{2}\cdot 2^x} = \frac{2^x}{2^x + \sqrt{2}} + \frac{2}{\sqrt{2}(\sqrt{2} + 2^x)} = \frac{2^x}{2^x + \sqrt{2}} + \frac{\sqrt{2}}{2^x + \sqrt{2}} = 1\,, $$ so that $$f(x) + f(1-x) = 1$$ for all $$x\,. $$

It follows that in the sum each term can be paired with its complement, namely $$ f\Bigl(\tfrac{k}{82}\Bigr) + f\Bigl(1 - \tfrac{k}{82}\Bigr) = f\Bigl(\tfrac{k}{82}\Bigr) + f\Bigl(\tfrac{82-k}{82}\Bigr) = 1. $$ Pairing $$k=1$$ with $$k=81$$, $$k=2$$ with $$k=80$$, …, $$k=40$$ with $$k=42$$ yields 40 pairs each summing to 1, while $$k=41$$ remains unpaired.

The remaining middle term is $$ f\bigl(\tfrac{41}{82}\bigr) = f\bigl(\tfrac12\bigr) = \frac{2^{1/2}}{2^{1/2} + \sqrt2} = \frac{\sqrt2}{\sqrt2 + \sqrt2} = \frac{1}{2}\,. $$ Therefore $$ \sum_{k=1}^{81} f\Bigl(\tfrac{k}{82}\Bigr) = 40\cdot 1 + \tfrac12 = \frac{81}{2}\,. $$

The correct answer is Option D: $$\frac{81}{2}$$.