Question

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

Answer 1

Answer: 4

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) $]$
$=\frac{{\alpha }^{n+1}+{\beta }^{n+1}+\alpha \beta {\alpha }^{n-1}+\alpha \beta {\beta }^{n-1}}{{\alpha }^n+{\beta }^n}=\frac{\left({\alpha }^n+{\beta }^n\right)(\alpha +\beta )}{\left({\alpha }^n+{\beta }^n\right)}=\alpha +\beta =4$.