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)
$\Rightarrow ({\alpha }^{13}-{\beta }^{13})-12({\alpha }^{12}-{\beta }^{12})=4({\alpha }^{11}-{\beta }^{11})$
$\Rightarrow t_{13}-12t_{12}=4t_{11}$
$(\because t_n={\alpha }^n-{\beta }^n)$
$\Rightarrow \frac{t_{13}-4t_{11}}{6t_{12}}=2$