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 $a=0=b.$

$a_{12}=$

• A. $a_{11}-a_{10}$
• B. $a_{11}+a_{10}$
• C. $2a_{11}+a_{10}$
• D. $a_{11}+2a_{10}$

Answer 1

The correct option is B $a_{11}+a_{10}$

$\alpha ,\beta$ are the roots of the equation $x^2-x-1=0$
Therefore, ${\alpha }^2-\alpha -1=0\Rightarrow {\alpha }^2=\alpha +1$
and ${\beta }^2-\beta -1=0\Rightarrow {\beta }^2=\beta +1$

$a_{n+1}+a_n=p{\alpha }^{n+1}+q{\beta }^{n+1}+p{\alpha }^n+q{\beta }^n\Rightarrow a_{n+1}+a_n=p{\alpha }^n(\alpha +1)+q{\beta }^n(\beta +1)=p{\alpha }^n{\alpha }^2+q{\beta }^n{\beta }^2=p{\alpha }^{n+2}+q{\beta }^{n+2}$
$\Rightarrow a_{n+1}+a_n=a_{n+2}$
$\therefore a_{11}+a_{10}=a_{12}$