f(x)=4^x/[(4^x)+2] = 1 / [1 + 2^(1-2x)]
f(1-x) = 1 / [1 + 2^(2x -1)] = 2^(1-2x) / [2^(1-2x) + 1]
→ f(x) + f(1-x) = 1
f(1/101)+f(2/101)+f(3/101)+...+f(100/101) = {f(1/101) + f(100/101)} + ...+ {f(50/101) + f(51/101)}