Question

The Olympiad task For the 10th grade if you don 't know how to solve it , say it right away . There is no time limit for a solution, the main thing is to solve in detail

Answer 1

Given

$\begin{gathered} f(\frac{x}{1+x})=\frac{1}{2}f(x) \\ f(1-x)=1-f(x) \end{gathered}$

Putting

$x=1$

In the first equation, we have,

$f(\frac{1}{2})=\frac{1}{2}f(1)$

Now, putting

$x=0$

in the first equation,

$\begin{gathered} f(0)=\frac{1}{2}f(0) \\ f(0)=0 \end{gathered}$

Putting x=0 in the second equation,

$\begin{gathered} f(1)=1-f(0) \\ =1 \end{gathered}$

So, the value of f(1) is 1. It implies

$f(\frac{1}{2})=\frac{1}{2}$

Now,

$\begin{gathered} f(\frac{1}{3})=f(1-\frac{2}{3}) \\ =1-f(\frac{2}{3}) \\ =1-f(\frac{2}{2+1}) \\ =1-\frac{1}{2}f(2) \end{gathered}$$\begin{gathered} f(\frac{1}{4})=f(1-\frac{3}{4}) \\ =1-f(\frac{3}{4}) \\ =1-f(\frac{3}{3+1}) \\ =1-\frac{1}{2}f(3) \end{gathered}$

Continuing in this process, we have,

$f(\frac{1}{n})=1-\frac{1}{2}f(n-1)$

Now, let us add the terms.

$\begin{gathered} f(1)+f(\frac{1}{2})+f(\frac{1}{3})+\cdots+f(\frac{1}{n}) \\ =1+1-\frac{1}{2}f(1)+1-\frac{1}{2}f(2)+\cdots+1-\frac{1}{2}f(n-1) \\ =(1+1+1+\cdots+1)-\frac{1}{2}\lbrack f(1)+f(2)+f(3)+\cdots+f(n-1)\rbrack \\ =n-\frac{1}{2}\lbrack f(1)+f(2)+f(3)+\cdots+f(n-1)\rbrack \\ =n-\frac{1}{2}\lbrack f(1)+1-f(-1)+1-f(-2)+\cdots+1-f(-n)\rbrack \\ =n-\frac{1}{2}\lbrack(1+1+\cdots2\text{n times)-4\rbrack} \\ =n-\frac{1}{2}\lbrack2n-4\rbrack \\ =n-n+2 \\ =2 \end{gathered}$

Hence, the sum is 2.