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

Question

Let $p, q$ be integers and let $\alpha, \beta$ be the roots of the equation, $x^{2}-x-1=0$, where $\alpha 
eq \beta$. For $n=0,1,2, \ldots$, let $a_{n}=p \alpha^{n}+q \beta^{n}$. FACT : If $a$ and $b$ are rational numbers and $a+b \sqrt{5}=0$, then $a=0=b$.\ $a_{12}=$\

Professor PC Thomas Classes · Answer

(d) : As $\alpha$ and $\beta$ are the roots, we have $\alpha^{2}=\alpha+1 ; \beta^{2}=\beta+1$\ Multiply by $p \alpha^{n}$ and $q \beta^{n}$ respectively, we have\ $p \alpha^{n+2}=p \alpha^{n+1}+p \alpha^{n}$\ $q \beta^{n+2}=q \beta^{n+1}+q \beta^{n}$\ Adding, we get\ $p \alpha^{n+2}+q \beta^{n+2}=\left(p \alpha^{n+1}+q \beta^{n+1}ight)+\left(p \alpha^{n}+q \beta^{n}ight)$ which becomes,\ $a_{n+2}=a_{n+1}+a_{n}, n \geq 0$\ Setting $n=10$, we have $a_{12}=a_{11}+a_{10}$\