Mathematics - Application of Derivative Question with Solution | TestHub

Since $\alpha ,\beta$ be the roots of the equation $x^2-ax+b=0$, so each of them must satisfy the equation. Therefore ${\alpha }^2-a\alpha +b=0\dots (1)$ and ${\beta }^2-a\beta +b=0$ $\dots (2)$ Now, $A_{n+1}-a{A}_n+b{A}_{n-1}={\alpha }^{n+1}+{\beta }^{n+1}$ $-a\left({\alpha }^n+{\beta }^n\right)+b\left({\alpha }^{n-1}+{\beta }^{n-1}\right)$ $={\alpha }^{n-1}\left({\alpha }^2-a\alpha +b\right)+{\beta }^{n-1}\left({\beta }^2-a\beta +b\right)$ $={\alpha }^{n-1}(0)+{\beta }^{n-1}(0)=0[$ From $(1)\&(2)]$