JEE Main 2022 — Mathematics Algebra

Given,

${P}_{n}={\alpha }^{n}-{\beta }^{n}$Pn​=αn−βn and \alpha &\beta are roots of ${x}^{2}-x-4=0$x2−x−4=0

Now putting $n-1$n−1 in ${P}_{n}={\alpha }^{n}-{\beta }^{n}$Pn​=αn−βn we get,

$${P}_{n-1}=({\alpha }^{n-1}-{\beta }^{n-1})$$Pn−1​=(αn−1−βn−1)

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

$${P}_{n}-{P}_{n-1}=({\alpha }^{n}-{\beta }^{n})-({\alpha }^{n-1}-{\beta }^{n-1})$$Pn​−Pn−1​=(αn−βn)−(αn−1−βn−1)

$$\Rightarrow {P}_{n}-{P}_{n-1}={\alpha }^{n-2}({\alpha }^{2}-\alpha )-{\beta }^{n-2}({\beta }^{2}-\beta )$$⇒Pn​−Pn−1​=αn−2(α2−α)−βn−2(β2−β)

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

$${P}_{n}-{P}_{n-1}=4({\alpha }^{n-2}-{\beta }^{n-2})$$Pn​−Pn−1​=4(αn−2−βn−2)

$${P}_{n}-{P}_{n-1}=4{P}_{n-2}$$Pn​−Pn−1​=4Pn−2​

Now putting the value in given expression

$$\frac{{P}_{15}{P}_{16}-{P}_{14}{P}_{16}-{P}_{15}^{2}+{P}_{14}{P}_{15}}{{P}_{13}{P}_{14}}$$P13​P14​P15​P16​−P14​P16​−P152​+P14​P15​​

$$=\frac{{P}_{16}({P}_{15}-{P}_{14})-{P}_{15}({P}_{15}-{P}_{14})}{{P}_{13}{P}_{14}}$$=P13​P14​P16​(P15​−P14​)−P15​(P15​−P14​)​

$$=\frac{({P}_{15}-{P}_{14})({P}_{16}-{P}_{15})}{{P}_{13}{P}_{14}}=\frac{(4{P}_{13})(4{P}_{14})}{{P}_{13}{P}_{14}}=16$$=P13​P14​(P15​−P14​)(P16​−P15​)​=P13​P14​(4P13​)(4P14​)​=16