While I was going through past Olympiad math papers, I found this question without any explanation. Here is the question:
Given a function $f(x)=\frac{9^{x}}{9^x+3}$, what is $f(\frac{1}{27})+ f(\frac{2}{27}) + f(\frac{3}{27})+ ...+ f(\frac{26}{27})$?
The answer was 13.
I took a really bad approach and converted $\frac{9^{x}}{9^x+3}$ to $1+\frac{9^{x}}{3}$, which I then noticed was wrong.
I also accidentally multiplied $9^{\frac{1}{27}}$ with $9^{\frac{2}{27}}$, $9^{\frac{2}{27}}$ with $9^{\frac{3}{27}}$, and so on, before realizing that the functions were added and not multiplied.
I suspect that there is something to the power of $\frac{n}{27}$, because 9 is a multiple of 27. However, I am not completely sure.
Is there a law that tells me how I can solve this question? Since this is a Math Olympiad question, there is probably a maximum time limit of five minutes to do this question. This means that I probably won’t have time for tedious mathematical calculations with a calculator and online tools, or something like that. Please give me a quick, fast solution that is probably suitable for an 8th grader, at most a solution at a 10th grader level.
See I'm a $10^{th} $ grader & this is my APPROACH :
Here we can see , $$f(x)+f(1-x)=1$$
So , I can say , $$\underbrace{\left(f(\frac{1}{27})+f(\frac{26}{27})\right)+.......+\left(f(\frac{13}{27})+f(\frac{14}{27})\right)}_{n=26}$$ OR [ Here $n$ represents number of Terms .] $$\underbrace{1+1+1+.....+1}_{n=13}$$
Hence the answer is $13$ .