QJEEM22Maf3e1: JEE Main 2022 Mathematics - Class 11 JEEM22Maf3e1 Solution: Let be the roots of the quadratic equation If , , then is equal …

$\alpha$ and $\beta$ are the roots of the quadratic equation ${x^2} - x - 4 = 0$ .

$\therefore$ $\alpha$ and $\beta$ are satisfy the given equation.

${\alpha ^2} - \alpha - 4 = 0$

$\Rightarrow {\alpha ^{n + 1}} - {\alpha ^n} - 4{\alpha ^{n - 1}} = 0$ ...... (1)

and ${\beta ^2} - \beta - 4 = 0$

$\Rightarrow {\beta ^{n + 1}} - {\beta ^n} - 4{\beta ^{n - 1}} = 0$ ...... (2)

Subtracting (2) from (1), we get,

$({\alpha ^{n + 1}} - {\beta ^{n + 1}}) - ({\alpha ^n} - {\beta ^n}) - 4({\alpha ^{n - 1}} - {\beta ^{n - 1}}) = 0$

$\Rightarrow {P_{n + 1}} - {P_n} - 4{P_{n - 1}} = 0$

$\Rightarrow {P_{n + 1}} = {P_n} + 4{P_{n - 1}}$

$\Rightarrow {P_{n + 1}} - {P_n} = 4{P_{n - 1}}$

For $n = 14$ , ${P_{15}} - {P_{14}} = 4{P_{13}}$

For $n = 15$ , ${P_{16}} - {P_{15}} = 4{P_{14}}$

Now, ${{{P_{15}}{P_{16}} - {P_{14}}{P_{16}} - P_{15}^2 + {P_{14}}{P_{15}}} \over {{P_{13}}{P_{14}}}}$

$= {{{P_{16}}({P_{15}} - {P_{14}}) - {P_{15}}({P_{15}} - {P_{14}})} \over {{P_{13}}{P_{14}}}}$

$= {{({P_{15}} - {P_{14}})({P_{16}} - {P_{15}})} \over {{P_{13}}{P_{14}}}}$

$= {{(4{P_{13}})(4{P_{14}})} \over {{P_{13}}{P_{14}}}}$

$= 16$