For $$x \in R$$, $$x \neq 0$$, $$x \neq 1$$, let $$f_0(x) = \frac{1}{1-x}$$ and $$f_{n+1}(x) = f_0(f_n(x))$$, $$n = 0, 1, 2, \ldots$$. Then the value of $$f_{100}(3) + f_1\left(\frac{2}{3}\right) + f_2\left(\frac{3}{2}\right)$$ is equal to:

We are given the function $$ f_0(x) = \frac{1}{1-x} $$ for $$ x \in \mathbb{R} $$, $$ x \neq 0 $$, $$ x \neq 1 $$, and the recursive definition $$ f_{n+1}(x) = f_0(f_n(x)) $$ for $$ n = 0, 1, 2, \ldots $$. We need to find the value of $$ f_{100}(3) + f_1\left(\frac{2}{3}\right) + f_2\left(\frac{3}{2}\right) $$.

First, we compute the first few functions to identify a pattern. Start with $$ f_0(x) $$:

$$ f_0(x) = \frac{1}{1-x} $$

Now, compute $$ f_1(x) = f_0(f_0(x)) $$:

$$ f_1(x) = f_0\left( \frac{1}{1-x} \right) = \frac{1}{1 - \frac{1}{1-x}} $$

Simplify the denominator:

$$ 1 - \frac{1}{1-x} = \frac{1-x}{1-x} - \frac{1}{1-x} = \frac{(1-x) - 1}{1-x} = \frac{-x}{1-x} $$

So,

$$ f_1(x) = \frac{1}{\frac{-x}{1-x}} = \frac{1-x}{-x} = -\frac{1-x}{x} = \frac{x-1}{x} $$

Next, compute $$ f_2(x) = f_0(f_1(x)) $$:

$$ f_2(x) = f_0\left( \frac{x-1}{x} \right) = \frac{1}{1 - \frac{x-1}{x}} $$

Simplify the denominator:

$$ 1 - \frac{x-1}{x} = \frac{x}{x} - \frac{x-1}{x} = \frac{x - (x-1)}{x} = \frac{1}{x} $$

So,

$$ f_2(x) = \frac{1}{\frac{1}{x}} = x $$

Now, compute $$ f_3(x) = f_0(f_2(x)) $$:

$$ f_3(x) = f_0(f_2(x)) = f_0(x) = \frac{1}{1-x} $$

This is the same as $$ f_0(x) $$. Similarly,

$$ f_4(x) = f_0(f_3(x)) = f_0\left( \frac{1}{1-x} \right) = f_1(x) = \frac{x-1}{x} $$

$$ f_5(x) = f_0(f_4(x)) = f_0\left( \frac{x-1}{x} \right) = f_2(x) = x $$

We observe that the functions repeat every 3 steps: $$ f_3 = f_0 $$, $$ f_4 = f_1 $$, $$ f_5 = f_2 $$, and so on. Thus, the sequence is periodic with period 3, meaning:

$$ f_n(x) = f_{n \mod 3}(x) $$

where:

• If $$ n \equiv 0 \pmod{3} $$, then $$ f_n(x) = f_0(x) = \frac{1}{1-x} $$
• If $$ n \equiv 1 \pmod{3} $$, then $$ f_n(x) = f_1(x) = \frac{x-1}{x} $$
• If $$ n \equiv 2 \pmod{3} $$, then $$ f_n(x) = f_2(x) = x $$

Now, we compute each term in the expression $$ f_{100}(3) + f_1\left(\frac{2}{3}\right) + f_2\left(\frac{3}{2}\right) $$.

First, for $$ f_{100}(3) $$:

Find $$ 100 \mod 3 $$: $$ 100 \div 3 = 33 \times 3 = 99 $$, remainder $$ 100 - 99 = 1 $$, so $$ 100 \equiv 1 \pmod{3} $$. Thus, $$ f_{100}(3) = f_1(3) $$.

Using $$ f_1(x) = \frac{x-1}{x} $$:

$$ f_1(3) = \frac{3-1}{3} = \frac{2}{3} $$

So, $$ f_{100}(3) = \frac{2}{3} $$.

Second, for $$ f_1\left(\frac{2}{3}\right) $$:

Using $$ f_1(x) = \frac{x-1}{x} $$:

$$ f_1\left(\frac{2}{3}\right) = \frac{\frac{2}{3} - 1}{\frac{2}{3}} = \frac{-\frac{1}{3}}{\frac{2}{3}} = -\frac{1}{3} \times \frac{3}{2} = -\frac{1}{2} $$

Third, for $$ f_2\left(\frac{3}{2}\right) $$:

Since $$ f_2(x) = x $$:

$$ f_2\left(\frac{3}{2}\right) = \frac{3}{2} $$

Now, sum the terms:

$$ f_{100}(3) + f_1\left(\frac{2}{3}\right) + f_2\left(\frac{3}{2}\right) = \frac{2}{3} + \left(-\frac{1}{2}\right) + \frac{3}{2} $$

Combine the terms step by step. First, combine the fractions with denominator 2:

$$ -\frac{1}{2} + \frac{3}{2} = \frac{-1 + 3}{2} = \frac{2}{2} = 1 $$

Then add to $$ \frac{2}{3} $$:

$$ \frac{2}{3} + 1 = \frac{2}{3} + \frac{3}{3} = \frac{5}{3} $$

Hence, the value is $$ \frac{5}{3} $$. Comparing with the options:

A. $$ \frac{8}{3} $$

B. $$ \frac{4}{3} $$

C. $$ \frac{5}{3} $$

D. $$ \frac{1}{3} $$

So, the correct answer is Option C.