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(B) Given that $\alpha$ and $\beta$ are roots of the quadratic equation $x^{2} - (\alpha+\beta)x + \alpha\beta = 0$.Substituting the values,we get $x^

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$324$

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(B) Given that $\alpha$ and $\beta$ are roots of the quadratic equation $x^{2} - (\alpha+\beta)x + \alpha\beta = 0$.Substituting the values,we get $x^{2} - x - 1 = 0$.Since $\alpha$ and $\beta$ are roots,they satisfy the equation:$\alpha^{2} - \alpha - 1 = 0 \Rightarrow \alpha^{n+1} = \alpha^{n} + \alpha^{n-1}$$\beta^{2} - \beta - 1 = 0 \Rightarrow \beta^{n+1} = \beta^{n} + \beta^{n-1}$Adding these two equations,we get:$(\alpha^{n+1} + \beta^{n+1}) = (\alpha^{n} + \beta^{n}) + (\alpha^{n-1} + \beta^{n-1})$This implies the recurrence relation $p_{n+1} = p_{n} + p_{n-1}$.Given $p_{n+1} = 29$ and $p_{n-1} = 11$,we have:$29 = p_{n} + 11$$p_{n} = 29 - 11 = 18$.Therefore,$p_{n}^{2} = 18^{2} = 324$.

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 .... .

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