Let `alpha and beta` are the roots of `x^2 – x – 1 = 0` such that `P_k = alpha^k + beta^k , k ge 1` then which one is incorrect?

Options

A. `(p_(1)+p_(2)+p_(3)+p_(4)+p_(5))=26`
B. `p_(3)=p_(5)-p_(4)`
C. `p_(5)=11`
D. `p_(5)=p_(2).P_(3)`

Answer

To solve the problem, we need to analyze the roots of the quadratic equation \(x^2 - x - 1 = 0\) and compute the values of \(P_k = \alpha^k + \beta^k\) for \(k = 1, 2, 3, 4, 5\).

### Step 1: Find the roots \(\alpha\) and \(\beta\)
The roots of the equation can be found using the quadratic formula:
\[ \alpha, \beta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For our equation \(x^2 - x - 1 = 0\), we have \(a = 1\), \(b = -1\), and \(c = -1\):
\[ \alpha, \beta = \frac{1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{1 \pm \sqrt{5}}{2} \]
Thus, the roots are:
\[ \alpha = \frac{1 + \sqrt{5}}{2}, \quad \beta = \frac{1 - \sqrt{5}}{2} \]

### Step 2: Calculate \(P_1\)
\[ P_1 = \alpha + \beta = 1 \quad (\text{sum of the roots}) \]

### Step 3: Calculate \(P_2\)
Using the identity \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\):
\[ P_2 = \alpha^2 + \beta^2 = (1)^2 - 2(-1) = 1 + 2 = 3 \]

### Step 4: Calculate \(P_3\)
Using the recurrence relation \(P_k = \alpha P_{k-1} + \beta P_{k-1}\):
\[ P_3 = \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 + \beta^2) - \alpha\beta = 1 \cdot 3 - (-1) = 3 + 1 = 4 \]

### Step 5: Calculate \(P_4\)
Using the recurrence relation:
\[ P_4 = \alpha P_3 + \beta P_3 = P_3 + P_2 = 4 + 3 = 7 \]

### Step 6: Calculate \(P_5\)
Using the recurrence relation:
\[ P_5 = \alpha P_4 + \beta P_4 = P_4 + P_3 = 7 + 4 = 11 \]

### Summary of values
- \(P_1 = 1\)
- \(P_2 = 3\)
- \(P_3 = 4\)
- \(P_4 = 7\)
- \(P_5 = 11\)

### Step 7: Analyze the options
1. The sum \(P_1 + P_2 + P_3 + P_4 + P_5 = 1 + 3 + 4 + 7 + 11 = 26\) (Correct)
2. \(P_5 = 11\) (Correct)
3. \(P_5 = 11\) (Correct)
4. \(P_5 \neq P_2 \cdot P_3\) since \(3 \cdot 4 = 12\) (Incorrect)

### Conclusion
The incorrect statement is:
\[ P_5 = P_2 \cdot P_3 \]