Prove the following
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

Solution:

Given, $\alpha+\beta=1, \alpha \beta=-1$

$\therefore$ Quadratic equation with roots $\alpha, \beta$ is $x^{2}-x-1=0 \Rightarrow \alpha^{2}=\alpha+1$

Multiplying both sides by $\alpha^{n-1} \alpha^{n+1}=\alpha^{n}+\alpha^{n-1}$ (1)

Similarly, $\beta^{n+1}=\beta^{n}+\beta^{n-1} \ldots(2)$

Adding (1)\&(2) $\alpha^{\mathrm{n}+1}+\beta^{\mathrm{n}+1}=\left(\alpha^{\mathrm{n}}+\beta^{\mathrm{n}}\right)+\left(\alpha^{\mathrm{n}-1}+\beta^{\mathrm{n}-1}\right)$

$\Rightarrow P_{n+1}=P_{n}+P_{n-1}$

$\Rightarrow 29=P_{n}+11\left(\right.$ Given $\left., P_{n+1}=29, P_{n-1}=11\right)$

$\Rightarrow P_{n}=18$

$\therefore P_{n}^{2}=18^{2}=324$