Calculate, $f\bigg(\frac{1}{1997}\bigg)+f\bigg(\frac{2}{1997}\bigg)+f\bigg(\frac{3}{1997}\bigg)\ldots f\bigg(\frac{1996}{1997}\bigg)$ 
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?

I would say your method is practically speaking what I would also do. 
Maybe I would rephrase it as follows: 
Claim: $f(a)+f(1-a)=1$. 
Then write $S$ for the sum in question, and then $2S$ can be written as $f(1/1997+1996/1997) + \cdots$  (the Gauss trick), which is $1996$ by the claim, so $S=998$. 
