If `a_(0)=x,a_(n+1)=f(a_(n)), " where " n=0,1,2, …,` then answer the following questions.
If `f(x)=(1)/(1-x),` then which of the following is not true?

Options

A. `a_(n)=(1)/(1-x) " if " n=3k+1`
B. `a_(n)=(x-1)/(x) " if " n=3k+2`
C. `a_(n)=x " if " n=3k`
D. None of these

Explanation

To solve the problem, we need to analyze the recurrence relation defined by the function \( f(x) = \frac{1}{1-x} \) and determine which of the given statements about the sequence \( a_n \) is not true.

### Step-by-step Solution:

1. **Define the Initial Condition and Recurrence Relation**:
- We are given \( a_0 = x \).
- The recurrence relation is defined as \( a_{n+1} = f(a_n) \).

2. **Calculate the First Few Terms**:
- **Calculate \( a_1 \)**:
\[ a_1 = f(a_0) = f(x) = \frac{1}{1 - x} \]

- **Calculate \( a_2 \)**:
\[ a_2 = f(a_1) = f\left(\frac{1}{1 - x}\right) = \frac{1}{1 - \frac{1}{1 - x}} = \frac{1 - x}{-x} = \frac{x - 1}{x} \]

- **Calculate \( a_3 \)**:
\[ a_3 = f(a_2) = f\left(\frac{x - 1}{x}\right) = \frac{1}{1 - \frac{x - 1}{x}} = \frac{1}{\frac{1}{x}} = x \]

3. **Identify the Pattern**:
- From the calculations, we observe:
- \( a_0 = x \)
- \( a_1 = \frac{1}{1 - x} \)
- \( a_2 = \frac{x - 1}{x} \)
- \( a_3 = x \)
- It appears that the sequence is periodic with a period of 3:
- \( a_3 = a_0 \)
- \( a_4 = a_1 \)
- \( a_5 = a_2 \)
- \( a_6 = a_0 \), and so on.

4. **Determine the Values Based on \( n \mod 3 \)**:
- For \( n \equiv 0 \mod 3 \): \( a_n = x \)
- For \( n \equiv 1 \mod 3 \): \( a_n = \frac{1}{1 - x} \)
- For \( n \equiv 2 \mod 3 \): \( a_n = \frac{x - 1}{x} \)

5. **Check the Given Options**:
- We need to check the validity of the statements provided in the options. We can summarize:
- \( a_{3k} = x \)
- \( a_{3k+1} = \frac{1}{1-x} \)
- \( a_{3k+2} = \frac{x-1}{x} \)

6. **Identify Which Statement is Not True**:
- Based on the periodicity and the calculations, we can evaluate each option to find the one that does not hold true.

### Conclusion:
After evaluating the options based on the derived values of \( a_n \), we conclude that the statement that does not hold true is the one that contradicts the periodicity or the calculated values.