Find the value of $E(\frac{1}{2018})+E(\frac{2}{2018})+\dots+E(\frac{2017}{2018})$ Let $E(n) = \dfrac{4^n}{4^n+2}$.
If the value of $E(\frac{1}{2018})+E(\frac{2}{2018})+\dots+E(\frac{2017}{2018})=\frac{a}{b}$ (in lowest terms), find $b$.
I tried solving this question using telescopic sum but was unable to solve

Let $\displaystyle f(x)=\frac{4^x}{4^x+2},$ Then $\displaystyle f(1-x)=\frac{4^{1-x}}{4^{1-x}+2}=\frac{2}{4^x+2}$
So $$f(x)+f(1-x)=\frac{4^x+2}{4^x+2}=1.$$
So Put $\displaystyle x=\frac{1}{2018},\frac{2}{2018},\frac{3}{2018},\cdots \cdots ,\frac{1008}{2018}$
So $$f\bigg(\frac{1}{2018}\bigg)+f\bigg(\frac{2}{2018}\bigg)+f\bigg(\frac{3}{2018}\bigg)+\cdots +\cdots +f\bigg(\frac{1009}{2018}\bigg)+f\bigg(\frac{1010}{2018}\bigg)\cdots +f\bigg(\frac{2017}{2018}\bigg)=1008+\frac{1}{2}$$
For calculation of $\displaystyle f\bigg(\frac{1009}{2018}\bigg),$ 
Put $\displaystyle x=\frac{1009}{2018}$ in $f(x)+f(1-x)=1,$ 
Getting $$\displaystyle f\bigg(\frac{1009}{2018}\bigg)=\frac{1}{2}.$$
