Let p - q be integers and - - be the roots of the equation x 2-2 x -3-0 where -If an-p -n-q -n where n -0-1-2- - - - then the value of a9 is A- 3 a 8-5 a 7-3 a 6B- a 7-6 a 6- a 7-12 a 5-18 a 4D- 2 a 8-3 a 6

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Roots of a Quadratic Equation

Let p , q be ...

Question

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

3 a_{8} - 5 a_{7} + 3 a_{6}

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a_{7} - 6 a_{6}

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a_{7} - 12 a_{5} + 18 a_{4}

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2 a_{8} + 3 a_{6}

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Solution

The correct options are
A $3 a_{8} - 5 a_{7} + 3 a_{6}$
B $a_{7} - 6 a_{6}$
C $a_{7} - 12 a_{5} + 18 a_{4}$
$a_{n + 2} - 2 a_{n + 1} - 3 a_{n} = p (α^{n + 2}) + q (β^{n + 2}) - 2 (p α^{n + 1} + q β^{n + 1}) + 3 (p α^{n} + q β^{n})$
$= p α^{n} (α^{2} - 2 α + 3) + q β^{n} (β^{2} - 2 β + 3) = 0$
$a_{n + 2} = 2 a_{n + 1} - 3 a_{n}$
$a_{9} = 2 a_{8} - 3 a_{7} \Rightarrow a_{9} = 3 a_{8} - a_{8} - 3 a_{7}$
$a_{8} = 2 a_{7} - 3 a_{6}$
$a_{9} = 3 a_{8} - (2 a_{7} - 3 a_{6}) - 3 a_{7}$
$a_{9} = 3 a_{8} - 5 a_{7} + 3 a_{6}$

$a_{9} = 2 a_{8} - 3 a_{7}$
$a_{9} = 2 (2 a_{7} - 3 a_{6}) - 3 a_{7}$
$(∵$ substituting $a_{7} = 2 a_{6} - 3 a_{5}$ in the above equation )
$a_{9} = a_{7} - 6 a_{6}$

substituting $a_{6} = 2 a_{5} - 3 a_{4}$ in the above equation
$a_{9} = a_{7} - 6 (2 a_{5} - 3 a_{4})$
$a_{9} = a_{7} - 12 a_{5} + 18 a_{4}$

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be integers and

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x^{2} - 2 x + 3 = 0

where

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