Question

If $\alpha$ and $\beta$ are roots of quadratic equation $x^2-4 x+1=0$ such that $a_n=\alpha^n+\beta^n$, then the find value of $\frac{a_{n+1}+a_{n-1}}{a_n}$.

Answer 1

Clearly, $\alpha+\beta=4, \alpha \beta=1$
Now, $\frac{a_{n+1}+a_{n-1}}{a_n}=\frac{\alpha^{n+1}+\beta^{n+1}+\alpha^{n-1}+\beta^{n-1}}{\alpha^n+\beta^n} [$ As, $\alpha \beta=1$ (given) $]$
$=\frac{\alpha^{n+1}+\beta^{n+1}+\alpha \beta \alpha^{n-1}+\alpha \beta \beta^{n-1}}{\alpha^n+\beta^n}=\frac{\left(\alpha^n+\beta^n\right)(\alpha+\beta)}{\left(\alpha^n+\beta^n\right)}=\alpha+\beta=4$.