Question

Let $\alpha$ and $\beta$ be two real numbers such that $\alpha+\beta=1$ and $\alpha \beta=-1 .$ Let $p _{ n }=(\alpha)^{ n } (\beta)^{ n }$ $p _{ n -1}=11$ and $p _{ n +1}=29$ for some integer $n \geq 1$. Then, the value of $p _{ n }^{2}$ is _______.

Answer 1

$x ^{2}- x -1=0 \quad$ roots $=\alpha, \beta$
$\alpha^{2}-\alpha-1=0 \Rightarrow \alpha^{ n +1}=\alpha^{ n }+\alpha^{ n -1}$
$\beta^{2}-\beta-1=0 \Rightarrow \beta^{ n +1}=\beta^{ n }+\beta^{ n -1}$
$\frac{+}{ P _{ n +1}= P _{ n }+ P _{ n -1}}$
$29= P _{ n }+11$
$P _{ n }=18$
$P _{ n }^{2}=324$