If $\mathrm{\alpha }, \mathrm{\beta }$ are the roots of ${\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}=0$ and ${\mathrm{S}}_{\mathrm{n}}={\mathrm{\alpha }}^{\mathrm{n}}+{\mathrm{\beta }}^{\mathrm{n}}$ then ${\mathrm{aS}}_{\mathrm{n}+1}+{\mathrm{bS}}_{\mathrm{n}}+{\mathrm{cS}}_{\mathrm{n}-1}=$

$\begin{array}{l}GivenEquationa{x}^2+bx+c=0\\ multiplyingwith{x}^{n-1}onbothsides\\ \Rightarrow a{x}^{n+1}+b{x}^n+c{x}^{n-1}=0\\ \alpha ,\beta arerootsoftheEquationthen\\ a{\alpha }^{n+1}+b{\alpha }^n+c{\alpha }^{n-1}=0--\left(1\right)\\ anda{\beta }^{n+1}+b{\beta }^n+c{\beta }^{n-1}=0--\left(2\right)\\ \left(1\right)+\left(2\right)a({\alpha }^{n+1}+{\beta }^{n+1})+b({\alpha }^n+{\beta }^n)+c({\alpha }^{n-1}+{\beta }^{n-1})=0\\ \therefore a{s}_{n+1}+b{s}_n+c{s}_{n-1}=0\end{array}$