Given:-
[tex]f(\frac{1}{x+1})=\frac{1}{2}f(x),f(1-x)=1-f(x)[/tex]To find:-
[tex]s_n=f(1)+f(\frac{1}{2})+f(\frac{1}{3})+\cdots+f(\frac{1}{n})[/tex]Now we simplify above equation to solve it,
[tex]\begin{gathered} f(1)+f(\frac{1}{2})+f(\frac{1}{3})+\cdots+f(\frac{1}{n})=f(1)+f(\frac{2-1}{2})+f(\frac{3-2}{3})+\cdots+f(\frac{n+1-n}{n}) \\ f(1)+f(\frac{1}{2})+f(\frac{1}{3})+\cdots+f(\frac{1}{n})=f(1)+f(1-\frac{1}{2})+f(1-\frac{2}{3})+\ldots+f(1-\frac{(n-1)}{n}) \\ f(1)+f(\frac{1}{2})+f(\frac{1}{3})+\cdots+f(\frac{1}{n})=f(1)+1-f(\frac{1}{2})+1-f(\frac{2}{3})+\cdots+1-f(\frac{n-1}{n}) \\ f(1)+f(\frac{1}{2})+f(\frac{1}{3})+\cdots+f(\frac{1}{n})=f(1)-f(\frac{1}{2})-f(\frac{2}{3})-\cdots-f(\frac{n-1}{n})+1(n-1) \\ f(1)+f(\frac{1}{2})+f(\frac{1}{3})+\cdots+f(\frac{1}{n})=f(1)-f(\frac{1}{1+1})-f(\frac{2}{2+1})-\cdots-f(\frac{n-1}{(n-1)+1})+1(n-1) \end{gathered}[/tex]By furthur simplification we get,
[tex]\begin{gathered} f(1)-f(\frac{1}{1+1})-f(\frac{2}{2+1})-\cdots-f(\frac{n-1}{(n-1)+1})+1(n-1)=f(1)-\frac{1}{2}f(1)-\frac{1}{2}f(2)-\cdots-\frac{1}{2}f(n-1)+1(n-1)_{} \\ \text{ =}-\frac{1}{2}f(1)-\frac{1}{2}f(2)-\cdots-\frac{1}{2}f(n-1)+1(n-1) \\ \text{ =-}\frac{1}{2}\lbrack f(1)+f(2)+...+f(n-1)\rbrack+1(n-1) \end{gathered}[/tex]So the requiired answer is,
[tex]\text{-}\frac{1}{2}\lbrack f(1)+f(2)+...+f(n-1)\rbrack+1(n-1)[/tex]