For, x∈R,x≠0,x≠1, let f0(x)=11−xand fn+1(x)=f0fn(x),n=0,1,2,….Then the value of f100(3)+f123+f232 is equal to :
f0(x)=11−x f1(x)=f011−x=x−1x f2(x)=x; f100(3)+f123+f232=23+23−123+32=53