For x ∈ R, x≠ 0, x≠ 1, let f_0(x)= 1 1-x and f_n+1(x)=f_0(f_n(x)), n=0, 1, 2,… .. Then the value of f_100(3)+f_1( 2 3)+f_2( 3 2) is equal to :

$${f}_{1}(x)={f}_{0+1}(x)={f}_{0}({f}_{0}(x))=\frac{1}{1-\frac{1}{1-x}}=\frac{x-1}{x}$$f1​(x)=f0+1​(x)=f0​(f0​(x))=1−1−x1​1​=xx−1​

$${f}_{2}(x)={f}_{1+1}(x)={f}_{0}({f}_{1}(x))=\frac{1}{1-\frac{x-1}{x}}=x$$f2​(x)=f1+1​(x)=f0​(f1​(x))=1−xx−1​1​=x

$${f}_{3}(x)={f}_{2+1}(x)={f}_{0}({f}_{2}(x))={f}_{0}(x)=\frac{1}{1-x}$$f3​(x)=f2+1​(x)=f0​(f2​(x))=f0​(x)=1−x1​

$${f}_{4}(x)={f}_{3+1}(x)={f}_{0}({f}_{3}(x))=\frac{x-1}{x}$$f4​(x)=f3+1​(x)=f0​(f3​(x))=xx−1​

$\therefore$∴ ${f}_{0}={f}_{3}={f}_{6}=\ldots =\frac{1}{1-x}$f0​=f3​=f6​=…=1−x1​

$${f}_{1}={f}_{4}={f}_{7}=\ldots =\frac{x-1}{x}$$f1​=f4​=f7​=…=xx−1​

$${f}_{2}={f}_{5}={f}_{8}=\ldots =x$$f2​=f5​=f8​=…=x

$${f}_{100}(3)=\frac{3-1}{3}=\frac{2}{3},$$f100​(3)=33−1​=32​,

$${f}_{1}(\frac{2}{3})=\frac{\frac{2}{3}-1}{\frac{2}{3}}= -\frac{1}{2}$$f1​(32​)=32​32​−1​=−21​

$${f}_{2}(\frac{3}{2})=\frac{3}{2}$$f2​(23​)=23​

$\therefore$∴ ${f}_{100}(3)+{f}_{1}(\frac{2}{3})+{f}_{2}(\frac{3}{2})=\frac{5}{3}$f100​(3)+f1​(32​)+f2​(23​)=35​