The solution to a summation $4^n/(4^n+2 )$ I found this problem on summations, and I'm not really sure how to solve it. Could someone give a hint as to how to do so?
Find the value of
$$\sum_{i=1}^{1000}f\left(\frac{i}{1000}\right),\qquad f(x) = \frac{4^x}{4^x+2}$$ It came on an exam where we couldn't use calculators, and it apparently is an integer answer, though Wolfram Alpha disagrees...(Even if it isn't, I would still like to know how to do it)

We have
$$ f(x)+f(1-x) = \frac{4^x}{4^x+2}+\frac{4^{1-x}}{4^{1-x}+2} = \frac{4^x}{4^x+2}+\frac{2}{4^x+2}=1 $$
hence
$$ \left[f\left(\frac{1}{1000}\right)+f\left(\frac{999}{1000}\right)\right]+\ldots+\left[f\left(\frac{499}{1000}\right)+f\left(\frac{501}{1000}\right)\right]=499 $$
and your sum is just $499+f(1)+f\left(\frac{1}{2}\right)=\color{red}{500+\frac{1}{6}}$.
