Let p- q be integers and - - be the roots of the equation x2 -2x -3 - 0 where -If an - p -n - q -n where n -0-1-2- then the value of a 9 is

A_n+2 2 a_n+1 +3a_n = 0 =p(α^n+2)+q β^n+2) 2 (p α^n+1+ q β^n+1) +3(p α^n + q β^n) nbsp; = p α^n (α^2 2 α +3) + qβ^n (β^2 2 β +3) = 0 a_n+2 =2a_n+1 3 a ...

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Byju's Answer

Standard XII

Mathematics

Roots of a Quadratic Equation

Let p, q be i...

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 option is C $a_{7} - 12 a_{5} + 18 a_{4}$
$a_{n + 2} - 2 a_{n + 1} + 3 a_{n} = 0 = p (α^{n + 2}) + q β^{n + 2}) - 2 (p α^{n + 1} + q β^{n + 1}) + 3 (p α^{n} + q β^{n})$

$= p α^{n} (α^{2} - 2 α + 3) + q β^{n} (β^{2} - 2 β + 3) = 0$

$a_{n + 2} = 2 a_{n + 1} - 3 a_{n}$
$a_{9} = 2 a_{8} - 3 a_{7} \Rightarrow a_{9} = 3 a_{8} - a_{8} - 3 a_{7}$

$a_{8} = 2 a_{7} - 3 a_{6}$
$a_{9} = 3 a_{8} - (2 a_{7} - 3 a_{6}) - 3 a_{7}$
$a_{9} = 3 a_{8} - 5 a_{7} + 3 a_{6}$

$a_{9} = 2 a_{8} - 3 a_{7}$
$a_{9} = 2 (2 a_{7} - 3 a_{6}) - 3 a_{7}$
$(∵$ 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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