Question

Let $\alpha ,\beta$ be the roots of $x^2-x-1=0$ $(\alpha >\beta )$ and $m,n\in Z,k\in W$ such that $a_k=m{\alpha }^k+n{\beta }^k.$ If $a_4=35,$ then the value of $3m+2n$ is equal to

• A. $85$
• B. $40$
• C. $25$
• D. $75$

Answer 1

The correct option is C $25$

$\alpha ,\beta$ are the roots of $x^2-x-1=0$
$\therefore \alpha ,\beta =\frac{1\pm \sqrt{5}}{2}\alpha =\frac{1+\sqrt{5}}{2},\beta =\frac{1-\sqrt{5}}{2}(\because \alpha >\beta )$

$a_k=m{\alpha }^k+n{\beta }^k\Rightarrow a_4=m{\alpha }^4+n{\beta }^4\cdots \left(1\right)$
Since, $\alpha$ is a root of $x^2-x-1=0$,
we have ${\alpha }^2=\alpha +1$
$\Rightarrow {\alpha }^4={\alpha }^2+2\alpha +1=3\alpha +2$
Similarly, ${\beta }^4=3\beta +2$
Substituting these values in equation $\left(1\right)$, we get
$35=m(3\alpha +2)+n(3\beta +2)\Rightarrow 35=m\left(\frac{7+3\sqrt{5}}{2}\right)+n\left(\frac{7-3\sqrt{5}}{2}\right)\Rightarrow 35=\frac{7}{2}(m+n)+\frac{3\sqrt{5}}{2}(m-n)$
Equating rational and irrational parts,
$\frac{7}{2}(m+n)=35$ and $\frac{3\sqrt{5}}{2}(m-n)=0$
$\Rightarrow m+n=10$ and $m-n=0$
$\Rightarrow m=n=5\therefore 3m+2n=25$