Question

Let $\alpha$ and $\beta$ be roots of $x^{2}-12x-4=0 (\alpha >\, \beta)$ and $t_{n}=\alpha^{n}-\beta^{n} \, \forall \, n \, \in\,N$, then the value of $\frac{t_{13}-4t_{11}}{6t_{12}}$ equals

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

Answer: (A)

$\alpha, \beta (\alpha >\, \beta)$ are roots of $x^{2}-12x-4=0$
$\therefore \alpha^{2}-12\alpha-4=0 \, \dots(i)$
and $\beta^{2}-12\beta-4=0 \, \dots(ii)$
$\Rightarrow \alpha^{13}-12\alpha^{12}-4 \alpha^{11}-0$ and $\beta^{13}-12\beta^{12}-4^{11}=0$
(On multiplying the above equations by $\alpha^{11}$ & $\beta^{11}$ respectively)
$\Rightarrow (\alpha^{13}-\beta^{13})-12 (\alpha^{12}-\beta^{12})=4(\alpha^{11}-\beta^{11})$
$\Rightarrow t_{13}-12t_{12}=4t_{11}$
$(\because t_{n}=\alpha^{n}-\beta^{n})$
$\Rightarrow \frac{t_{13}-4t_{11}}{6t_{12}}=2$