3a. For $x>0$ let $f(x)=4^{x} /\left(4^{x}+2\right)$. Determine the value of the sum

$$
\sum_{k=1}^{1290} f\left(\frac{k}{1291}\right)
$$

## Solution
The following applies for $0<x<1 / 2$

$$
f\left(\frac{1}{2}+x\right)+f\left(\frac{1}{2}-x\right)=\frac{2 \cdot 4^{x}}{2 \cdot 4^{x}+2}+\frac{2 \cdot 4^{-x}}{2 \cdot 4^{-x}+2}=\frac{4^{x}}{4^{x}+1}+\frac{1}{1+4^{x}}=1
$$

where we have truncated both fractions by 2 and expanded the second one by $4^{x}$. This now easily results in

$$
\sum_{k=1}^{1290} f\left(\frac{k}{1291}\right)=\sum_{k=1}^{645} f\left(\frac{k}{1291}\right)+f\left(1-\frac{k}{1291}\right)=\sum_{k=1}^{645} 1=645
$$