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 ∈ ℕ , then P 15 P 16 - P 14 P 16 - P 15 2 + P 14 P 15 P 13 P 14 is equal to _____.

Question

Let &#945; , &#946; &#945; &#62; &#946; be the roots of the quadratic equation x 2 - x - 4 = 0 . If P n = &#945; n - &#946; n , n &#8712; &#8469; , then P 15 P 16 - P 14 P 16 - P 15 2 + P 14 P 15 P 13 P 14 is equal to _____.

Aicharya · Accepted Answer

Given, $P_n = \alpha^n - \beta^n$ and $\alpha$ & $\beta$ are roots of $x^2 - x - 4 = 0$ Now putting $n - 1$ in $P_n = \alpha^n - \beta^n$ we get, $P_{n-1} = (\alpha^{n-1} - \beta^{n-1})$ Now subtracting $P_n - P_{n-1}$ we get, $P_n - P_{n-1} = (\alpha^n - \beta^n) - (\alpha^{n-1} - \beta^{n-1})$ $\Rightarrow P_n - P_{n-1} = \alpha^{n-2}(\alpha^2 - \alpha) - \beta^{n-2}(\beta^2 - \beta)$ Now using the equation $\alpha^2 - \alpha - 4 = 0$ & $\beta^2 - \beta - 4 = 0$ we get, $P_n - P_{n-1} = 4(\alpha^{n-2} - \beta^{n-2})$ $P_n - P_{n-1} = 4P_{n-2}$ Now putting the value in given expression $\frac{P_{15}P_{16} - P_{14}P_{16} - P_{15}^2 + P_{14}P_{15}}{P_{13}P_{14}}$ $= \frac{P_{16}(P_{15} - P_{14}) - P_{15}(P_{15} - P_{14})}{P_{13}P_{14}}$ $= \frac{(P_{15} - P_{14})(P_{16} - P_{15})}{P_{13}P_{14}} = \frac{4P_{13} \cdot 4P_{14}}{P_{13}P_{14}} = 16$

Let α , β α > β be the roots of the quadratic equation x 2 - x - 4 = 0 . If P n =…

Solution

Let &#945; , &#946; &#945; &#62; &#946; be the roots of the quadratic equation x 2 - x - 4 = 0 . If P n =…

Solution

Let α , β α > β be the roots of the quadratic equation x 2 - x - 4 = 0 . If P n =…