Question

If $\alpha$ and $\beta$ are the roots of equation $x^2-3x+1=0$ and $a_n={\alpha }^n+{\beta }^n,n\in N$ then the value of $\frac{a_7+a_5}{a_6}$

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is C $3$

$x^2-3x+1=0\to \alpha ,\beta$ are roots $\alpha \beta =1,\alpha +\beta =3$.

$a_n={\alpha }^n+{\beta }^n.$

$\frac{a_7+a_5}{a_6}=\frac{{\alpha }^7+{\beta }^7+{\alpha }^5+{\beta }^5}{{\alpha }^6+{\beta }^6}$

$=\frac{\alpha }{1+{\left(\frac{\beta }{\beta }\right)}^6}+\frac{\beta }{(1+{\left(\frac{\beta }{\alpha }\right)}^6)}\frac{1}{\alpha \left(1{\left(\frac{\beta }{\alpha }\right)}^6\right)}+\frac{1}{\beta (1+{\left(\frac{\alpha }{\beta }\right)}^6)}$

$\frac{\alpha }{\beta }=\frac{\beta }{\alpha }=1$

$=\frac{\alpha }{2}+\frac{\beta }{2}+\frac{1}{2\alpha }+\frac{1}{2\beta }$

$=\frac{3}{2}+\frac{\alpha +\beta }{2\alpha \beta }$

$=\frac{3}{2}+\frac{3}{2}$

$=3$

$C$ is correct.