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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 .... .
$162$
$324$
$648$
$424$
Solution
$x ^{2}- x -1=0 \quad$ roots $=\alpha, \beta$
$\alpha^{2}-\alpha-1=0 \Rightarrow \alpha^{ n +1}=\alpha^{ n }+\alpha^{ n -1}$
$\beta^{2}-\beta-1=0 \Rightarrow \beta^{ n +1}=\beta^{ n }+\beta^{ n -1}$
$\quad\quad\quad\quad\quad\quad\quad\quad+$_______________________
$\quad\quad\quad\quad\quad\quad\quad\quad P_{n+1}=P_{n}+P_{n-1}$
$\quad\quad\quad\quad\quad\quad\quad\quad 29=P_{n}+11$
$\quad\quad\quad\quad\quad\quad\quad\quad P_{n}=18$
$\quad\quad\quad\quad\quad\quad\quad\quad P_{n}^{2}=324$
