If $f(x) = \frac{4^x}{4^x+2},$ find the value of $\sum \limits_{i=1}^{1999} f\left(\frac{i}{1999}\right) $

Solution 1:

Hint:

Observe

$$f(x)+f(1-x)=1$$

So, $$\begin{align}&f\left(\dfrac 1 {1999}\right)+f\left(\dfrac 2 {1999}\right)+ \cdots+f\left(\dfrac{1998}{1999}\right)+f\left(\dfrac{1999}{1999}\right)\\&=f\left(\dfrac 1 {1999}\right)+f\left(\dfrac{1998}{1999}\right)+f\left(\dfrac 2 {1999}\right)+f\left(\dfrac {1997} {1999}\right)+\cdots +f\left(\dfrac{999}{1999}\right)+f\left(\dfrac{1000}{1999}\right)+f(1)\\&=999+f(1)\\&=999+\dfrac 2 3\\&=\dfrac {2999} 3\end{align}$$

$$\begin{align}f(x)+f(1-x)&=\dfrac{4^x}{4^x+2}+\dfrac{4^{1-x}}{4^{1-x}+2}\\&=\dfrac{4^x}{4^x+2}+\dfrac{\dfrac{4}{4^x}}{\dfrac{4}{4^x}+2}\\&=\dfrac{4^x}{4^x+2}+\dfrac 4 {2 \cdot4^x+4}\\&=\dfrac{4^x}{4^x+2}+\dfrac{\not{4}~~2}{\not2(4^x+2)}\\&=\dfrac{4^x+2}{4^x+2}\\&=1\end{align}$$