# Let $\alpha\; \text{and} \; \beta$ be two real numbers such that $\alpha + \beta =1 \text{ and } \alpha \beta =-1$. Let $p_n =(\alpha )^n+(\beta )^n$, $p_{n-1} =11 \text{ and } p_{n+1}=29$ for some integer $n\geqslant 1$. Then, the value of $p^{2}_{n}$ is ________ Option: 1 324 Option: 2 264 Option: 3 128 Option: 4 64

Given that

$\\\alpha+\beta=1\;\;\text{and}\;\;\alpha\beta=-1\\\text{So, the quadratic equation is }\\\Rightarrow x^2-x-1=0\\\Rightarrow\alpha^{2}-\alpha-1=0\\\text{Multiply by }\alpha^{n-1}\\\Rightarrow\alpha^{n+1}=\alpha+\alpha^{n-1}\\\text{similarly,}\\\Rightarrow\beta^{n+1}=\beta+\beta^{n-1}$

$\\\alpha^{n+1}+\beta^{n+1}=\alpha^n+\beta^n+\alpha^{n-1}+\beta^{n-1}\\\Rightarrow P_{n+1}=P-n+P_{n-1}$

$\\29=P_{n}+11 \\ P_{n}=18 \\ P_{n}^{2}=324$

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