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((x)/(1+x))=(1)/(2)f(x) \\ f(1-x)=1-f(x) \end{gathered}`

Putting

`x=1`

In the first equation, we have,

`f((1)/(2))=(1)/(2)f(1)`

Now, putting

`x=0`

in the first equation,

`\begin{gathered} f(0)=(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((1)/(2))=(1)/(2)`

Now,

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

Continuing in this process, we have,

`f((1)/(n))=1-(1)/(2)f(n-1)`

Now, let us add the terms.

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

Hence, the sum is 2.