Question

Let $p, q$ be integers and $a, \beta$ be the roots of the equation $x^{2} - 2x + 3 = 0$ where $a \neq \beta$. If $a_{n} = p a^{n} + q \beta^{n}$ where $n \in {0, 1, 2, \ldots}$, then the value of $a_{9}$ is:

(A) $3a_{8} - 5a_{7} + 3a_{6}$

(B) $a_{7} - 6a_{6}$

(C) $a_{7} - 12a_{5} + 18a_{4}$

(D) $2a_{8} + 3a_{6}$.

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Answer

The correct options are:

  • (A) $3 a_{8}-5 a_{7}+3 a_{6}$
  • (B) $a_{7}-6 a_{6}$
  • (C) $a_{7}-12 a_{5}+18 a_{4}$

Given the sequence $a_n = p a^n + q \beta^n$ where $a$ and $\beta$ are the roots of the quadratic equation $x^2 - 2x + 3 = 0$, we derive the following:

The quadratic $x^2 - 2x + 3 = 0$ factors into: $$ a^2 - 2a + 3 = 0 \quad \text{and} \quad \beta^2 - 2\beta + 3 = 0 $$

Thus, we can write: $$ \alpha^2 - 2 \alpha + 3 = 0 \quad \text{and} \quad \beta^2 - 2 \beta + 3 = 0 $$

Now, consider the sequence relation: $$ a_{n+2} - 2a_{n+1} + 3a_n = p \alpha^{n+2} + q \beta^{n+2} - 2(p a^{n+1} + q \beta^{n+1}) + 3(p a^n + q \beta^n) = 0 $$ Therefore, $$ a_{n+2} = 2a_{n+1} - 3a_n $$

Using the recurrence relation, we need to find the expression for $a_9$.

First, we use: $$ a_{9} = 2a_{8} - 3a_{7} $$

To derive further: $$ a_{8} = 2a_{7} - 3a_{6} $$

Thus, $$ a_{9} = 2(2a_{7} - 3a_{6}) - 3a_{7} \ = 4a_{7} - 6a_{6} - 3a_{7} \ = a_{7} - 6a_{6} $$

Similarly, for $a_9$ derived from substituting $a_{6}$: $$ a_{7} = 2a_{6} - 3a_{5} $$

So, $$ a_{9} = a_{7} - 6(2a_{5} - 3a_{4}) \ = a_{7} - 12a_{5} + 18a_{4} $$

Finally, providing the option that matches, $$ a_{9} = 3a_{8} - 5a_{7} + 3a_{6} $$

Therefore, the answers are:

  • (A) $3 a_{8}-5 a_{7}+3 a_{6}$
  • (B) $a_{7}-6 a_{6}$
  • (C) $a_{7}-12 a_{5}+18 a_{4}$

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