Let $\alpha, \beta(\alpha>\beta)$ be the roots of the quadratic equation $x^{2}-x-4=0 .$ If $P_{n}=\alpha^{n}-\beta^{n}$, $n \in \mathrm{N}$, 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 (integer)
16
Solution
<p>$\alpha$ and $\beta$ are the roots of the quadratic equation ${x^2} - x - 4 = 0$.</p>
<p>$\therefore$ $\alpha$ and $\beta$ are satisfy the given equation.</p>
<p>${\alpha ^2} - \alpha - 4 = 0$</p>
<p>$\Rightarrow {\alpha ^{n + 1}} - {\alpha ^n} - 4{\alpha ^{n - 1}} = 0$ ...... (1)</p>
<p>and ${\beta ^2} - \beta - 4 = 0$</p>
<p>$\Rightarrow {\beta ^{n + 1}} - {\beta ^n} - 4{\beta ^{n - 1}} = 0$ ...... (2)</p>
<p> Subtracting (2) from (1), we get,</p>
<p>$$({\alpha ^{n + 1}} - {\beta ^{n + 1}}) - ({\alpha ^n} - {\beta ^n}) - 4({\alpha ^{n - 1}} - {\beta ^{n - 1}}) = 0$$</p>
<p>$\Rightarrow {P_{n + 1}} - {P_n} - 4{P_{n - 1}} = 0$</p>
<p>$\Rightarrow {P_{n + 1}} = {P_n} + 4{P_{n - 1}}$</p>
<p>$\Rightarrow {P_{n + 1}} - {P_n} = 4{P_{n - 1}}$</p>
<p>For $n = 14$, ${P_{15}} - {P_{14}} = 4{P_{13}}$</p>
<p>For $n = 15$, ${P_{16}} - {P_{15}} = 4{P_{14}}$</p>
<p>Now, $${{{P_{15}}{P_{16}} - {P_{14}}{P_{16}} - P_{15}^2 + {P_{14}}{P_{15}}} \over {{P_{13}}{P_{14}}}}$$</p>
<p>$$ = {{{P_{16}}({P_{15}} - {P_{14}}) - {P_{15}}({P_{15}} - {P_{14}})} \over {{P_{13}}{P_{14}}}}$$</p>
<p>$= {{({P_{15}} - {P_{14}})({P_{16}} - {P_{15}})} \over {{P_{13}}{P_{14}}}}$</p>
<p>$= {{(4{P_{13}})(4{P_{14}})} \over {{P_{13}}{P_{14}}}}$</p>
<p>$= 16$</p>
About this question
Subject: Mathematics · Chapter: Complex Numbers and Quadratic Equations · Topic: Complex Numbers and Argand Plane
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