Let α, β (α > β) be the roots of the quadratic equation x ^ 2 - x -...

Given,

${P}_{n} = {\alpha}^{n} - {\beta}^{n}$ and $\alpha \text{ and } \beta$ are roots of ${x}^{2} - x - 4 = 0$

Now putting $n - 1$ in ${P}_{n} = {\alpha}^{n} - {\beta}^{n}$ we get,

${P}_{n - 1} = \left({\alpha}^{n - 1} - {\beta}^{n - 1}\right)$

Now subtracting ${P}_{n} - {P}_{n - 1}$ we get,

${P}_{n} - {P}_{n - 1} = \left({\alpha}^{n} - {\beta}^{n}\right) - \left({\alpha}^{n - 1} - {\beta}^{n - 1}\right)$

$⇒ {P}_{n} - {P}_{n - 1} = {\alpha}^{n - 2}\left({\alpha}^{2} - \alpha\right) - {\beta}^{n - 2}\left({\beta}^{2} - \beta\right)$

Now using the equation ${\alpha}^{2} - \alpha - 4 = 0 \text{ and } {\beta}^{2} - \beta - 4 = 0$ we get,

${P}_{n} - {P}_{n - 1} = 4\left({\alpha}^{n - 2} - {\beta}^{n - 2}\right)$

${P}_{n} - {P}_{n - 1} = 4{P}_{n - 2}$

Now putting the value in given expression

$\dfrac{{P}_{15}{P}_{16} - {P}_{14}{P}_{16} - {P}_{15}^{2} + {P}_{14}{P}_{15}}{{P}_{13}{P}_{14}}$

$= \dfrac{{P}_{16}\left({P}_{15} - {P}_{14}\right) - {P}_{15}\left({P}_{15} - {P}_{14}\right)}{{P}_{13}{P}_{14}}$

$= \dfrac{\left({P}_{15} - {P}_{14}\right)\left({P}_{16} - {P}_{15}\right)}{{P}_{13}{P}_{14}} = \dfrac{\left(4{P}_{13}\right)\left(4{P}_{14}\right)}{{P}_{13}{P}_{14}} = 16$