TITLE: Calculate, $f\bigg(\frac{1}{1997}\bigg)+f\bigg(\frac{2}{1997}\bigg)+f\bigg(\frac{3}{1997}\bigg)\ldots f\bigg(\frac{1996}{1997}\bigg)$
QUESTION [8 upvotes]: If $$f(x)=\frac{4^x}{4^x+2}$$
Calculate,
$$f\bigg(\frac{1}{1997}\bigg)+f\bigg(\frac{2}{1997}\bigg)+f\bigg(\frac{3}{1997}\bigg)\ldots
 f\bigg(\frac{1996}{1997}\bigg)$$

My Attempt:
I was not able to generalise the expression or get a solid pattern, so I started with smaller numbers and calculated,
$$f\bigg(\frac{1}{2}\bigg)=\frac{1}{2}$$
$$f\bigg(\frac{1}{3}\bigg)+f\bigg(\frac{2}{3}\bigg)=1$$
$$f\bigg(\frac{1}{4}\bigg)+f\bigg(\frac{2}{4}\bigg)+f\bigg(\frac{3}{4}\bigg)=\frac{3}{2}$$
I could see that,
$$f\bigg(\frac{1}{n}\bigg)+f\bigg(\frac{2}{n}\bigg)+f\bigg(\frac{3}{n}\bigg)\ldots
 f\bigg(\frac{n-1}{n}\bigg)=\frac{n-1}{2}$$
So,  $$f\bigg(\frac{1}{1997}\bigg)+f\bigg(\frac{2}{1997}\bigg)+f\bigg(\frac{3}{1997}\bigg)\ldots
 f\bigg(\frac{1996}{1997}\bigg)=998$$
which is indeed the right answer. But I am not satisfied with my method. How else can I solve it?

REPLY [4 votes]: Given $f(x)=\dfrac{4^x}{4^x+2}$
From that we get $f(1-x)=\dfrac{2}{4^x+2}$
First let us take the last term $f\left(\dfrac{1}{1997}\right)$
Notice that $f\left(\dfrac{1}{1997}\right)=f\left(1-\dfrac{1}{1997}\right)$ and same for the rest of the terms.
Now, $f(x)+f(1-x)=1$
$f\left(\dfrac{1}{1997}\right)+f\left(\dfrac{2}{1997}\right)+........+f\left(1-\dfrac{2}{1997}+f\left(1-\dfrac{1}{1997}\right)\right)$ 
all of them makes pairs.
So, the total pairs $=\dfrac{1996}{2}=998$
So, the sum $=1+1+1+......998$ times $=998$