`a_n=p(alpha^n)+q(beta^n) `where `alpha=(1+sqrt(5))/2 and beta = (1-sqrt(5))/2` , also `a_(n+1)=a_n+a_(n-1)`. If `a_(4) = 28, "then" p + 2q =`

Options

A. 21
B. 11
C. 7
D. 12

Explanation

To solve the problem step by step, we need to find the values of \( p \) and \( q \) given the expression for \( a_n \) and the recurrence relation.

### Step 1: Define the sequences
We start with the expression for \( a_n \):
\[ a_n = p \alpha^n + q \beta^n \]
where \( \alpha = \frac{1 + \sqrt{5}}{2} \) and \( \beta = \frac{1 - \sqrt{5}}{2} \).

### Step 2: Calculate \( a_0 \) and \( a_1 \)
Substituting \( n = 0 \):
\[ a_0 = p \alpha^0 + q \beta^0 = p + q \]

Substituting \( n = 1 \):
\[ a_1 = p \alpha^1 + q \beta^1 = p \alpha + q \beta \]

### Step 3: Use the recurrence relation to find \( a_2 \), \( a_3 \), and \( a_4 \)
Using the recurrence relation \( a_{n+1} = a_n + a_{n-1} \):

For \( a_2 \):
\[ a_2 = a_1 + a_0 = (p \alpha + q \beta) + (p + q) = p \alpha + q \beta + p + q \]

For \( a_3 \):
\[ a_3 = a_2 + a_1 = (p \alpha + q \beta + p + q) + (p \alpha + q \beta) = 2p \alpha + 2q \beta + p + q \]

For \( a_4 \):
\[ a_4 = a_3 + a_2 = (2p \alpha + 2q \beta + p + q) + (p \alpha + q \beta + p + q) \]
\[ = 3p \alpha + 3q \beta + 2(p + q) \]

### Step 4: Substitute the known value of \( a_4 \)
We know \( a_4 = 28 \):
\[ 28 = 3p \alpha + 3q \beta + 2(p + q) \]

### Step 5: Substitute the values of \( \alpha \) and \( \beta \)
Substituting \( \alpha \) and \( \beta \):
\[ 28 = 3p \left(\frac{1 + \sqrt{5}}{2}\right) + 3q \left(\frac{1 - \sqrt{5}}{2}\right) + 2(p + q) \]
\[ = \frac{3p(1 + \sqrt{5}) + 3q(1 - \sqrt{5})}{2} + 2(p + q) \]
\[ = \frac{3p + 3q + 3p \sqrt{5} - 3q \sqrt{5}}{2} + 2(p + q) \]

### Step 6: Combine like terms
Combine the terms:
\[ = \frac{3(p + q) + (3p - 3q) \sqrt{5}}{2} + 2(p + q) \]
\[ = \frac{3(p + q)}{2} + 2(p + q) + \frac{(3p - 3q) \sqrt{5}}{2} \]
\[ = \frac{3(p + q) + 4(p + q)}{2} + \frac{(3p - 3q) \sqrt{5}}{2} \]
\[ = \frac{7(p + q)}{2} + \frac{(3p - 3q) \sqrt{5}}{2} \]

### Step 7: Set up equations
Setting the rational and irrational parts equal:
1. \( \frac{7(p + q)}{2} = 28 \)
2. \( \frac{3(p - q) \sqrt{5}}{2} = 0 \)

From the second equation, since the coefficient of \( \sqrt{5} \) must equal zero:
\[ p - q = 0 \implies p = q \]

From the first equation:
\[ 7(p + q) = 56 \implies p + q = 8 \]
Since \( p = q \):
\[ 2p = 8 \implies p = 4, \quad q = 4 \]

### Step 8: Find \( p + 2q \)
Now we can find \( p + 2q \):
\[ p + 2q = 4 + 2 \times 4 = 4 + 8 = 12 \]

Thus, the final answer is:
\[ \boxed{12} \]