Mathematics - Quadratic Equation Question with Solution | TestHub

Given,

$P_n={\alpha }^{\mathrm{n}}-{\beta }^{\mathrm{n}}$ and $\alpha \& \beta$ are roots of $x^2-x-4=0$

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

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

Now subtracting $P_n-P_{n-1}$ we get,

$P_n-P_{n-1}=\left({\alpha }^n-{\beta }^n\right)-\left({\alpha }^{n-1}-{\beta }^{n-1}\right)$

$\Rightarrow P_n-P_{n-1}={\alpha }^{n-2}\left({\alpha }^2-\alpha \right)-{\beta }^{n-2}\left({\beta }^2-\beta \right)$

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

$P_n-P_{n-1}=4\left({\alpha }^{\mathrm{n}-2}-{\beta }^{\mathrm{n}-2}\right)$

$P_n-P_{n-1}=4P_{n-2}$

Now putting the value in given expression 

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

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

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