If $f(x)=\frac{2^{x}}{2^{x}+\sqrt{2}}, x \in R$, then $\sum_{k=1}^{81} f\left(\frac{k}{82}\right)$ i

Question

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

Professor PC Thomas Classes · Answer

(a): We have, $f(x)=\frac{2^{x}}{2^{x}+\sqrt{2}}, x \in R$\ $\Rightarrow f(x)+f(1-x)=\frac{2^{x}}{2^{x}+\sqrt{2}}+\frac{2^{1-x}}{2^{1-x}+\sqrt{2}}$\ $=\frac{2^{x}}{2^{x}+\sqrt{2}}+\frac{2}{2+\sqrt{2} \cdot 2^{x}}=1$\ Now, $S=\sum_{k=1}^{81} f\left(\frac{k}{82}ight)=f\left(\frac{1}{82}ight)+f\left(\frac{2}{82}ight)$ $$ +\ldots+f\left(\frac{81}{82}ight) $$ $\Rightarrow S=f\left(\frac{1}{82}ight)+\ldots+f\left(1-\frac{2}{82}ight)+f\left(1-\frac{1}{82}ight)$\ $=f\left(\frac{1}{82}ight)+f\left(1-\frac{1}{82}ight)+f\left(\frac{2}{82}ight)+f\left(1-\frac{2}{82}ight)$ $$ +\ldots+	ext { upto } 40+f\left(\frac{41}{82}ight) $$ $=\underbrace{(1+1+1+\ldots+1)}_{40 	ext { times }}+\frac{2^{1 / 2}}{2^{1 / 2}+\sqrt{2}}=40+\frac{1}{2}=\frac{81}{2}$\