If `x+(1)/(x)=-2` then the value of `x^(2n+1)+(1)/(x^(2n+1))` where n is a positive integer is

Options

A. `0`
B. `2`
C. `-2`
D. `-5`

Answer

To solve the problem, we need to find the value of \( x^{2n+1} + \frac{1}{x^{2n+1}} \) given that \( x + \frac{1}{x} = -2 \).

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ x + \frac{1}{x} = -2 \]

2. **Square both sides:**
\[ \left( x + \frac{1}{x} \right)^2 = (-2)^2 \]
This expands to:
\[ x^2 + 2 + \frac{1}{x^2} = 4 \]

3. **Rearranging the equation:**
\[ x^2 + \frac{1}{x^2} + 2 = 4 \]
Subtracting 2 from both sides gives:
\[ x^2 + \frac{1}{x^2} = 2 \]

4. **Now we need to find \( x^{2n+1} + \frac{1}{x^{2n+1}} \).** We can use the identity:
\[ x^{2n+1} + \frac{1}{x^{2n+1}} = \left( x^{2n-1} + \frac{1}{x^{2n-1}} \right) \left( x^2 + \frac{1}{x^2} \right) - \left( x^{2n-2} + \frac{1}{x^{2n-2}} \right) \]

5. **Base case for \( n=1 \):**
\[ x^{2(1)} + \frac{1}{x^{2(1)}} = x^2 + \frac{1}{x^2} = 2 \]

6. **Inductive step:**
For \( n=2 \):
\[ x^{2(2)+1} + \frac{1}{x^{2(2)+1}} = (x^{2(1)+1} + \frac{1}{x^{2(1)+1}})(x^2 + \frac{1}{x^2}) - (x^{2(0)+1} + \frac{1}{x^{2(0)+1}}) \]
Continuing this pattern, we find that:
\[ x^{2n+1} + \frac{1}{x^{2n+1}} = 2 \]
for all positive integers \( n \).

7. **Conclusion:**
Therefore, the value of \( x^{2n+1} + \frac{1}{x^{2n+1}} \) is:
\[ \boxed{2} \]