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(D) Given $P_{n} = \alpha^{n} - \beta^{n}$ and the quadratic equation $x^{2} - x - 4 = 0$.Since $\alpha$ and $\beta$ are roots,we have $\alpha^{2} = \

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$16$

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(D) Given $P_{n} = \alpha^{n} - \beta^{n}$ and the quadratic equation $x^{2} - x - 4 = 0$.Since $\alpha$ and $\beta$ are roots,we have $\alpha^{2} = \alpha + 4$ and $\beta^{2} = \beta + 4$.Consider $P_{n} - P_{n-1} = (\alpha^{n} - \beta^{n}) - (\alpha^{n-1} - \beta^{n-1}) = \alpha^{n-1}(\alpha - 1) - \beta^{n-1}(\beta - 1)$.Since $\alpha^{2} - \alpha = 4$ and $\beta^{2} - \beta = 4$,we have $\alpha - 1 = \frac{4}{\alpha}$ and $\beta - 1 = \frac{4}{\beta}$.Thus,$P_{n} - P_{n-1} = \alpha^{n-1} \cdot \frac{4}{\alpha} - \beta^{n-1} \cdot \frac{4}{\beta} = 4(\alpha^{n-2} - \beta^{n-2}) = 4P_{n-2}$.Now,simplify the expression: $\frac{P_{16}(P_{15} - P_{14}) - P_{15}(P_{15} - P_{14})}{P_{13} P_{14}} = \frac{(P_{15} - P_{14})(P_{16} - P_{15})}{P_{13} P_{14}}$.Using the relation $P_{n} - P_{n-1} = 4P_{n-2}$,we get $P_{15} - P_{14} = 4P_{13}$ and $P_{16} - P_{15} = 4P_{14}$.Substituting these values: $\frac{(4P_{13})(4P_{14})}{P_{13} P_{14}} = 16$.

Let $\alpha, \beta (\alpha > \beta)$ be the roots of the quadratic equation $x^{2} - x - 4 = 0$. If $P_{n} = \alpha^{n} - \beta^{n}, n \in N$,then $\frac{P_{15} P_{16} - P_{14} P_{16} - P_{15}^{2} + P_{14} P_{15}}{P_{13} P_{14}}$ is equal to $......$

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