If `a_(n+1)=(1)/(1-a_(n))" for " n ge 1 and a_(3)=a_(1)," then "(a_(2022))^(2022)` equals

Options

A. `-1`
B. 1
C. 0
D. none of these

Answer

To solve the problem step by step, we start with the given recurrence relation and the condition provided.

### Step 1: Understand the recurrence relation
We have the recurrence relation:
\[ a_{n+1} = \frac{1}{1 - a_n} \]
for \( n \geq 1 \) and the condition \( a_3 = a_1 \).

### Step 2: Find expressions for \( a_2 \) and \( a_3 \)
- For \( n = 1 \):
\[ a_2 = \frac{1}{1 - a_1} \]

- For \( n = 2 \):
\[ a_3 = \frac{1}{1 - a_2} \]

Since \( a_3 = a_1 \), we can substitute \( a_2 \) into this equation:
\[ a_1 = \frac{1}{1 - a_2} \]

### Step 3: Substitute \( a_2 \) into the equation for \( a_3 \)
Substituting \( a_2 = \frac{1}{1 - a_1} \) into the equation for \( a_3 \):
\[ a_1 = \frac{1}{1 - \left(\frac{1}{1 - a_1}\right)} \]

### Step 4: Simplify the equation
To simplify:
\[ a_1 = \frac{1}{1 - \frac{1}{1 - a_1}} = \frac{1}{\frac{(1 - a_1) - 1}{1 - a_1}} = \frac{1 - a_1}{-a_1} = \frac{a_1 - 1}{a_1} \]

Cross-multiplying gives:
\[ a_1^2 = a_1 - 1 \]
Rearranging this gives us the quadratic equation:
\[ a_1^2 - a_1 + 1 = 0 \]

### Step 5: Solve the quadratic equation
Using the quadratic formula \( a = 1, b = -1, c = 1 \):
\[ a_1 = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{1 \pm \sqrt{1 - 4}}{2} = \frac{1 \pm \sqrt{-3}}{2} \]
This gives:
\[ a_1 = \frac{1 \pm i\sqrt{3}}{2} \]

### Step 6: Identify the roots
Let:
\[ \omega = \frac{1 + i\sqrt{3}}{2}, \quad \omega^2 = \frac{1 - i\sqrt{3}}{2} \]
These are the cube roots of unity.

### Step 7: Find \( a_2 \)
Using the relation \( a_2 = \frac{1}{1 - a_1} \):
\[ a_2 = \frac{1}{1 - \omega} = \frac{1}{\omega^2} \]
Thus:
\[ a_2 = \omega \quad \text{or} \quad a_2 = \omega^2 \]

### Step 8: Determine the values of \( a_n \)
From the recurrence relation, we can see that:
- All odd indexed terms are equal to \( a_1 \) (which is \( \omega \) or \( \omega^2 \)).
- All even indexed terms are equal to \( a_2 \) (which is \( \omega^2 \) or \( \omega \)).

### Step 9: Find \( a_{2022} \)
Since \( 2022 \) is even:
\[ a_{2022} = a_2 = \omega \quad \text{or} \quad a_2 = \omega^2 \]

### Step 10: Calculate \( (a_{2022})^{2022} \)
Using \( a_{2022} = \omega \):
\[ (a_{2022})^{2022} = \omega^{2022} \]
Since \( \omega^3 = 1 \):
\[ 2022 \mod 3 = 0 \quad \Rightarrow \quad \omega^{2022} = (\omega^3)^{674} = 1^{674} = 1 \]

### Final Answer
Thus, the value of \( (a_{2022})^{2022} \) is:
\[ \boxed{1} \]