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

JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Sequences and Series

Let

α, β

(

α > β

) be the roots of the quadratic equation

x^{2} - x - 4 = 0

. If

P_{n} = α^{n} - β^{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:

Answer: 16

From the recurrence relation for the characteristic equation

x^{2} - x - 4 = 0

, we have

α^{2} = α + 4

and

β^{2} = β + 4

. This gives

P_{n} - P_{n - 1} = α^{n - 2} (α^{2} - α) - β^{n - 2} (β^{2} - β) = 4 (α^{n - 2} - β^{n - 2}) = 4 P_{n - 2}

. The expression becomes

\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}} = \frac{( 4 P _{13} ) ( 4 P _{14} )}{P _{13} P _{14}} = 16

JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Integer answer