Question

Let $p,q$ be integers and let $\alpha ,\beta$ be the roots of the equation, $x^2-x-1=0,$ where $\alpha \ne \beta .$ For $n=0,1,2,....,$ let $a_n=p{\alpha }^n+q{\beta }^n.$

FACT: If $a$ and $b$ are rational numbers and $a+b\sqrt{5}=0.$ then $a=0=b.$

If $a_4=28,$ then $p+2q=$

• A. $21$
• B. $14$
• C. $7$
• D. $12$

Answer 1

The correct option is D $12$

$\alpha ,\beta$ are the roots of the equation $x^2-x-1=0$
$\Rightarrow \alpha +\beta =1\Rightarrow \alpha =1-\beta \cdots \left(1\right)$
Also, ${\alpha }^2-\alpha -1=0\Rightarrow {\alpha }^2=\alpha +1$
$\Rightarrow {\alpha }^4=(\alpha +1{)}^2={\alpha }^2+2\alpha +1=3\alpha +2$
$\text{Similarly,}{\beta }^4=3\beta +2$

Now, $a_4=p{\alpha }^4+q{\beta }^4=28$
$\Rightarrow p(3\alpha +2)+q(3\beta +2)=28$
$\Rightarrow p(3\alpha +2)+q(5-3\alpha )=28\left[\text{From}\right(1\left)\right]\Rightarrow \alpha (3p-3q)+2p+5q=28$
$\Rightarrow p=q$ and $2p+5q=28\left[\text{Using the given fact}\right]$
$\Rightarrow p=q=4$
$\therefore p+2q=4+2\times 4=12$