Let P_n = alpha^n + beta^n, n in N. If P_{10} | vedclass

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(B) Given $P_n = \alpha^n + \beta^n$. We observe that $P_{10} = P_9 + P_8$ $(123 = 76 + 47)$.This implies that $\alpha$ and $\beta$ are roots of the q

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$x^2 + x - 1 = 0$

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(B) Given $P_n = \alpha^n + \beta^n$. We observe that $P_{10} = P_9 + P_8$ $(123 = 76 + 47)$.This implies that $\alpha$ and $\beta$ are roots of the quadratic equation $x^2 - x - 1 = 0$.For this equation,the sum of roots $\alpha + \beta = 1$ and the product of roots $\alpha \beta = -1$.We need to find the quadratic equation with roots $\frac{1}{\alpha}$ and $\frac{1}{\beta}$.The sum of the new roots is $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta} = \frac{1}{-1} = -1$.The product of the new roots is $\frac{1}{\alpha} \cdot \frac{1}{\beta} = \frac{1}{\alpha \beta} = \frac{1}{-1} = -1$.The required quadratic equation is $x^2 - (	ext{sum of roots})x + (	ext{product of roots}) = 0$.Substituting the values,we get $x^2 - (-1)x + (-1) = 0$,which simplifies to $x^2 + x - 1 = 0$.

Let $P_n = \alpha^n + \beta^n, n \in N$. If $P_{10} = 123, P_9 = 76, P_8 = 47$ and $P_1 = 1$,then the quadratic equation having roots $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ is:

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