Question

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

• A. $0$
• B. $abc$
• C. $a+b+c$
• D. $Noneofthese$

Answer 1

The correct option is A

$0$

Explanation for correct option:

Step 1: Find the sum and product of the roots of the given equation.

We have given a quadratic equation$a{x}^2+bx+c=0$, whose roots are$\alpha$ and$\beta$, $S_n={\alpha }^n+{\beta }^n$

we have to find the value of$a{S}_{n+1}+b{S}_n+c{S}_{n-1}$

So,

$$\begin{gathered} \alpha +\beta =\frac{-b}{a} \\ and\alpha \beta =\frac{c}{a} \end{gathered}$$

Step 2: Substitute the values of$\mathit{n}\mathbf{}\mathit{i}\mathit{n}\mathbf{}{\mathit{S}}_{\mathbf{n}}$.

$\begin{array}{rcl}{S}_{n+1}& =& {\alpha }^{n+1}+{\beta }^{n+1}\\ S_n& =& {\alpha }^n+{\beta }^n\\ S_{n-1}& =& {\alpha }^{n-1}+{\beta }^{n-1}\end{array}$

Step 3: Substituting the above values in the given expression$\mathit{a}{\mathit{S}}_{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{+}\mathit{b}{\mathit{S}}_{\mathbf{n}}\mathbf{+}\mathit{c}{\mathit{S}}_{\mathbf{n}\mathbf{-}\mathbf{1}}$

$$\begin{gathered} a{S}_{n+1}+b{S}_n+c{S}_{n-1}=a\{{\alpha }^{n+1}+{\beta }^{n+1}+\frac{b}{a}({\alpha }^n+{\beta }^n)+\frac{c}{a}({\alpha }^{n-1}+{\beta }^{n-1}\left)\right\} \\ \Rightarrow a\{{\alpha }^{n+1}+{\beta }^{n+1}-(\alpha +\beta \left)\right({\alpha }^n+{\beta }^n)+\alpha \beta ({\alpha }^{n-1}+{\beta }^{n-1}\left)\right\} \\ \Rightarrow a\{{\alpha }^{n+1}+{\beta }^{n+1}-{\alpha }^{n+1}-\alpha {\beta }^n-{\alpha }^n\beta -{\beta }^{n+1}+{\alpha }^n\beta +\alpha {\beta }^n\} \\ \Rightarrow a\left\{0\right\}=0 \\ \therefore a{S}_{n+1}+b{S}_n+c{S}_{n-1}=0 \end{gathered}$$

Hence, the option$\left(A\right)$ is correct.