Let $\alpha$ and $\beta$ be two real numbers such that $\alpha+\beta=1$ and $\alpha \beta=-1$. Let $

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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

Professor PC Thomas Classes · Answer

(324) : Quadratic equation whose roots are $\alpha, \beta$ is given by\ $x^{2}-x-1=0$\ $	herefore \quad \alpha^{2}-\alpha-1=0$ $$\begin{equation*}\Rightarrow \quad \alpha^{n+1}=\alpha^{n}+\alpha^{n-1} 	ag{i} \end{equation*}$$ and $\beta^{2}-\beta-1=0$ $$\begin{equation*}\Rightarrow \quad \beta^{n+1}=\beta^{n}+\beta^{n-1} 	ag{ii} \end{equation*}$$ Adding (i) and (ii), we get\ $\left(\alpha^{n+1}+\beta^{n+1}ight)=\left(\alpha^{n}+\beta^{n}ight)+\left(\alpha^{n-1}+\beta^{n-1}ight)$\ $\Rightarrow \quad p_{n+1}=p_{n}+p_{n-1} \Rightarrow 29=p_{n}+11$\ $\Rightarrow p_{n}=18 	herefore p_{n}{ }^{2}=324$\