If `u_n = 2cos n theta` then `u_1u_n - u_(n-1)` is equal to

Options

A. `u_(n +2)`
B. `u_(n+1)`
C. `u_2 u_(n+1)`
D. none of these

Answer

To solve the problem, we need to find the expression for \( u_1 u_n - u_{n-1} \) given that \( u_n = 2 \cos(n \theta) \).

### Step-by-Step Solution:

1. **Identify \( u_1 \), \( u_n \), and \( u_{n-1} \)**:
\[ u_1 = 2 \cos(\theta) \]
\[ u_n = 2 \cos(n \theta) \]
\[ u_{n-1} = 2 \cos((n-1) \theta) \]

2. **Substitute these values into the expression**:
We need to compute \( u_1 u_n - u_{n-1} \):
\[ u_1 u_n = (2 \cos(\theta))(2 \cos(n \theta)) = 4 \cos(\theta) \cos(n \theta) \]
Therefore,
\[ u_1 u_n - u_{n-1} = 4 \cos(\theta) \cos(n \theta) - 2 \cos((n-1) \theta) \]

3. **Use the product-to-sum identities**:
We can use the identity:
\[ 2 \cos A \cos B = \cos(A + B) + \cos(A - B) \]
Applying this identity to \( 4 \cos(\theta) \cos(n \theta) \):
\[ 4 \cos(\theta) \cos(n \theta) = 2 \cdot 2 \cos(\theta) \cos(n \theta) = 2(\cos(\theta + n \theta) + \cos(\theta - n \theta)) \]
This simplifies to:
\[ 2(\cos((n+1)\theta) + \cos((n-1)\theta)) \]

4. **Combine the terms**:
Now we substitute this back into our expression:
\[ u_1 u_n - u_{n-1} = 2(\cos((n+1)\theta) + \cos((n-1)\theta)) - 2 \cos((n-1)\theta) \]
The \( 2 \cos((n-1)\theta) \) terms cancel out:
\[ = 2 \cos((n+1)\theta) \]

5. **Relate it back to \( u_{n+1} \)**:
We know that:
\[ u_{n+1} = 2 \cos((n+1) \theta) \]
Thus, we can conclude:
\[ u_1 u_n - u_{n-1} = u_{n+1} \]

### Final Result:
\[ u_1 u_n - u_{n-1} = u_{n+1} \]