Question

lf $\alpha ,\beta$ are the roots of $a{x}^2+bx+c=0$ and $S_n={\alpha }^n+{\beta }^n$, then $a{S}_{n+1}+b{S}_n+c{S}_{n-1}=$

• A. $0$
• B. $1$
• C. $2$
• D. $5$

Answer 1

The correct option is A $0$

$Multiplyingboththesidesofthegivenequationby{x}^{n-1},wehavea{x}^{n+1}+b{x}^n+c{x}^{n-1}=0\because \alpha isaroot,theequationbecomesa{\alpha }^{n+1}+b{\alpha }^n+c{\alpha }^{n-1}=0.....\left(1\right)Similarly,taking\beta asarootofthegivenequation,wehave,a{\beta }^{n+1}+b{\beta }^n+c{\beta }^{n-1}=0.....\left(2\right)Addingequations\left(1\right)and\left(2\right)weget,a({\alpha }^{n+1}+{\beta }^{n+1})+b({\alpha }^n+{\beta }^n)+c({\alpha }^{n-1}+{\beta }^{n-1})=0\Rightarrow a{S}_{n+1}+b{S}_n+c{S}_{n-1}=0AnsOptionA.$