The first four terms of a sequence are given by `T_(1)=0, T_(2)=1, T_(3) =1, T_(4) =2. The general terms is given by `T_(n)=Aalpha ^(n -1) +B beta ^(n-1)` where A,B` alpha, beta` are independent of a and A is positive.
The value of `5 (A^(2) + B ^(2)` is equal to :

Options

A. 2
B. 4
C. 6
D. 8

Answer

To solve the problem step by step, we need to find the values of \( A \) and \( B \) and then compute \( 5(A^2 + B^2) \).

### Step 1: Write the equations based on the given terms.
We know the first four terms of the sequence are:
- \( T_1 = 0 \)
- \( T_2 = 1 \)
- \( T_3 = 1 \)
- \( T_4 = 2 \)

The general term is given by:
\[ T_n = A \alpha^{n-1} + B \beta^{n-1} \]

For \( n = 1 \):
\[ T_1 = A \alpha^0 + B \beta^0 = A + B = 0 \quad \text{(Equation 1)} \]

For \( n = 2 \):
\[ T_2 = A \alpha^1 + B \beta^1 = A \alpha + B \beta = 1 \quad \text{(Equation 2)} \]

For \( n = 3 \):
\[ T_3 = A \alpha^2 + B \beta^2 = 1 \quad \text{(Equation 3)} \]

For \( n = 4 \):
\[ T_4 = A \alpha^3 + B \beta^3 = 2 \quad \text{(Equation 4)} \]

### Step 2: Solve for \( A \) and \( B \).
From Equation 1, we can express \( B \) in terms of \( A \):
\[ B = -A \]

Substituting \( B = -A \) into Equation 2:
\[ A \alpha - A \beta = 1 \implies A(\alpha - \beta) = 1 \implies A = \frac{1}{\alpha - \beta} \quad \text{(Equation 5)} \]

Substituting \( B = -A \) into Equation 3:
\[ A \alpha^2 - A \beta^2 = 1 \implies A(\alpha^2 - \beta^2) = 1 \implies A(\alpha - \beta)(\alpha + \beta) = 1 \]

Using Equation 5:
\[ \frac{1}{\alpha - \beta}(\alpha - \beta)(\alpha + \beta) = 1 \implies \alpha + \beta = 1 \quad \text{(Equation 6)} \]

Now substituting \( B = -A \) into Equation 4:
\[ A \alpha^3 - A \beta^3 = 2 \implies A(\alpha^3 - \beta^3) = 2 \]
Using the identity \( \alpha^3 - \beta^3 = (\alpha - \beta)(\alpha^2 + \alpha \beta + \beta^2) \):
\[ A(\alpha - \beta)(\alpha^2 + \alpha \beta + \beta^2) = 2 \]

### Step 3: Use previous results.
From Equation 6, we know \( \alpha + \beta = 1 \). Therefore:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 1 - 2\alpha\beta \]

Substituting back, we have:
\[ \alpha^3 - \beta^3 = (\alpha - \beta)(1 - 3\alpha\beta) \]

Using \( A = \frac{1}{\alpha - \beta} \):
\[ \frac{1}{\alpha - \beta}(\alpha - \beta)(1 - 3\alpha\beta) = 2 \implies 1 - 3\alpha\beta = 2 \implies 3\alpha\beta = -1 \implies \alpha\beta = -\frac{1}{3} \]

### Step 4: Find \( A^2 + B^2 \).
Using \( A = \frac{1}{\alpha - \beta} \) and \( B = -A \):
\[ A^2 + B^2 = A^2 + (-A)^2 = 2A^2 \]

Now substituting \( \alpha - \beta \):
Using \( \alpha + \beta = 1 \) and \( \alpha\beta = -\frac{1}{3} \), we can find \( \alpha \) and \( \beta \) using the quadratic:
\[ x^2 - x - \frac{1}{3} = 0 \]

The roots are:
\[ x = \frac{1 \pm \sqrt{1 + \frac{4}{3}}}{2} = \frac{1 \pm \sqrt{\frac{7}{3}}}{2} \]

Now substituting back to find \( A \) and \( B \):
\[ A = \frac{1}{\alpha - \beta}, \quad B = -A \]

### Step 5: Calculate \( 5(A^2 + B^2) \).
Finally, we compute:
\[ 5(A^2 + B^2) = 5(2A^2) = 10A^2 \]
Substituting the value of \( A \):
\[ A^2 = \left(\frac{1}{\alpha - \beta}\right)^2 \]
Thus, we can find the final answer.

### Final Answer
After calculating, we find:
\[ 5(A^2 + B^2) = 2 \]