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

<p>$$\begin{aligned} & f(x)=\frac{2^x}{2^x+\sqrt{2}} \\ & f(x)+f(1-x)=\frac{2^x}{2^x+\sqrt{2}}+\frac{2^{1-x}}{2^{1-x}+\sqrt{2}} \\ & =\frac{2^x}{2^x+\sqrt{2}}+\frac{2}{2+\sqrt{2} 2^x}=\frac{2^x+\sqrt{2}}{2^x+\sqrt{2}}=1 \end{aligned}$$</p> <p>$$\begin{array}{r} \text { Now, } \sum_{\mathrm{k}=1}^{81} \mathrm{f}\left(\frac{\mathrm{k}}{82}\right)=\mathrm{f}\left(\frac{1}{82}\right)+\mathrm{f}\left(\frac{2}{82}\right)+\ldots \ldots+\mathrm{f}\left(\frac{81}{82}\right) \\ =\mathrm{f}\left(\frac{1}{82}\right)+\mathrm{f}\left(\frac{2}{82}\right)+\ldots \ldots .+\mathrm{f}\left(1-\frac{2}{82}\right)+\mathrm{f}\left(1-\frac{1}{82}\right) \\ =\left[\mathrm{f}\left(\frac{1}{82}\right)+\mathrm{f}\left(1-\frac{1}{82}\right)\right]+\left[\mathrm{f}\left(\frac{2}{82}\right)+\mathrm{f}\left(1-\frac{2}{82}\right)\right]+\ldots . .40 \text { cases }+\mathrm{f}\left(\frac{41}{82}\right) \end{array}$$</p> <p>$$\begin{aligned} & =(1+1+\ldots 40 \text { times })+\frac{2^{1 / 2}}{2^{1 / 2}+2^{1 / 2}} \\ & 40+\frac{1}{2}=\frac{81}{2} \end{aligned}$$</p>