Evaluating $\sum_{r=1}^{2001} f\left(\frac{r}{2002}\right)$, where $f(x)=\frac{4^{x}}{4^{x}+2}$ 
Let
  $
f(x)=\dfrac{4^{x}}{4^{x}+2}
$. Find
  $
\sum_{r=1}^{2001} f\left(\frac{r}{2002}\right)
$.

Given,
$$f(x)=\frac{4^{x}}{4^{x}+2}$$
$$
\begin{align}
\sum_{r=1}^{2001} f\left(\frac{r}{2002}\right) &=\sum_{r=1}^{2001} \frac{4^{\frac{r}{2002}}}{4^{\frac{r}{2002}}+2} \\[4pt]
&=\frac{4^{\sum_{r=1}^{2001}\frac{r}{2002}}}{4^{\sum_{r=1}^{2001}\frac{r}{2002}}+2}  \\[4pt]
&=\frac{4^{\frac{1}{2002}+\frac{2}{2002}+\cdots \cdot \frac{2001}{2002}}}{4^{\frac{1}{2002}+\frac{2}{2002}+\cdots \cdot \frac{2001}{2002}} +2}
\end{align}$$
Upon simplifying, I am getting
$$\frac{4^{{2001}}}{4^{{2001}}+2}$$

What to do next?

 A: There is a different way to solve this question.
You need use this identity $\mathit f(x) + f(1-x) = 1$. 
Let me prove this first $\pmb :$
$\mathit f(x) = \frac{4^x}{(4^x+2)}----->1 $
Replace $\mathit x\;$ as $\mathit 1-x\\$
$\mathit f(x) = \frac{4^{1-x}}{4^{1-x}+2} = \frac{4\cdot{4^{-x}}}{4\cdot{4^{-x}}+2}$
On simplifying you get $\mathit f(x) = \frac{4}{4+2\cdot{4^x}}$ $\;$ which simplifies as $\mathit = \;\frac{2}{2+{4^x}}----->2$
Add $\mathit equation \;\pmb1$ and $\pmb2\;$ you get the number $\pmb 1$ as the answer.
Now applying this in the summation $$\sum_{i=0}^{2001} \mathit f\left(\frac{r}{2002}\right) $$
Substituting values from $\pmb1$ to $\pmb{2001}$
$\mathit f\left(\frac{1}{2002}\right)$ + $\mathit f\left(\frac{2}{2002}\right)$ + ....... +  $\mathit f\left(\frac{1001}{2002}\right)$ + ....... $\mathit f\left(\frac{2000}{2002}\right)$ + $\mathit f\left(\frac{2001}{2002}\right)$
Group the terms $\pmb :$ $\mathit f\left(\frac{1}{2002}\right)$ + $\mathit f\left(\frac{2001}{2002}\right)$ + $\mathit f\left(\frac{2}{2002}\right)$ + $\mathit f\left(\frac{2000}{2002}\right)$ + ........ + $\mathit f\left(\frac{1000}{2002}\right)$ + $\mathit f\left(\frac{1002}{2002}\right)$ + $\mathit f\left(\frac{1001}{2002}\right)$
Now we have grouped $\pmb {2000}$ terms whose sum is $\pmb {1,}$ added $\pmb {2000}$ times. 
Finally we have to add the middle term $\mathit f\left(\frac{1001}{2002}\right)$ whose value we can find by substituting $\pmb {\frac{1}{2}}$ in the question $\mathit f(x) = \frac{4^x}{(4^x+2)}$. It's value is $\pmb {\frac{1}{2}}$.
So the final answer is $\mathit 1000 + \frac{1}{2}$ which is $\pmb {\frac{2001}{2}}$
Cheers!!
