Functions Part 2

$\;f(x)=\;\frac{2^x}{2^x+\sqrt{2}}$Consider the complementary argument:$\;f(1-x)=\;\frac{2^{1-x}}{2^{1-x}+\sqrt{2}}$Multiply numerator and denominator of $f(1-x)$ by $2^{x}$:$\;f(1-x)=\;\frac{2^{1-x}\cdot 2^{x}}{2^{1-x}\cdot 2^{x}+\sqrt{2}\cdot 2^{x}}=\;\frac{2}{2+\sqrt{2}\,2^{x}}$Also multiply numerator and denominator of $f(x)$ by $\sqrt{2}$:$\;f(x)=\;\frac{\sqrt{2}\,2^{x}}{\sqrt{2}\,2^{x}+2}=\;\frac{\sqrt{2}\,2^{x}}{2+\sqrt{2}\,2^{x}}$Adding the two results:$\;\Rightarrow f(x)+f(1-x)=\;\frac{2+\sqrt{2}\,2^{x}}{2+\sqrt{2}\,2^{x}}=1$So $f(x)+f(1-x)=1$ for every $x\in\mathbb{R}$.Now pair the terms that add up to $1$, i.e. $\frac{k}{82}+\frac{82-k}{82}=1$:$\;f\!\left(\frac{1}{82}\right)+f\!\left(\frac{81}{82}\right)=1,\;\; f\!\left(\frac{2}{82}\right)+f\!\left(\frac{80}{82}\right)=1,\;\;\dots,\;\; f\!\left(\frac{40}{82}\right)+f\!\left(\frac{42}{82}\right)=1$There are $40$ such pairs, and each pair contributes $1$, giving $40\times 1=40$.The only unpaired term is the middle one:$\;f\!\left(\frac{41}{82}\right)=f\!\left(\frac{1}{2}\right)=\;\frac{2^{1/2}}{2^{1/2}+\sqrt{2}}=\;\frac{\sqrt{2}}{\sqrt{2}+\sqrt{2}}=\;\frac{1}{2}$Therefore:$\;\sum\limits_{k=1}^{81} f\!\left(\frac{k}{82}\right)=40+\;\frac{1}{2}=\;\frac{81}{2}$Hence the correct option is (2), $\;\frac{81}{2}$.