Question

If $\alpha$ and $\beta$ are the roots of the equation $x^2-ax+b=0$ and$A_n={\alpha }^n+{\beta }^n$. Then $A_{n+1}-a{A}_n+b{A}_{n-1}$is equal to

• A. $-a$
• B. $b$
• C. $0$
• D. $a-b$

Answer 1

The correct option is C

$0$

Explanation for the correct option:

Find the value of ${\mathit{A}}_{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{}\mathbf{-}\mathbf{}\mathit{a}{\mathit{A}}_{\mathbf{n}}\mathbf{}\mathbf{+}\mathbf{}\mathit{b}{\mathit{A}}_{\mathbf{n}\mathbf{-}\mathbf{1}}$:

If $\alpha$ and $\beta$ are the roots of the equation $x^2-ax+b=0$, so each of them must satisfy the equation.

$\Rightarrow {\alpha }^2–a\alpha +b=0\dots ....\left(1\right)$ and ${\beta }^2–a\beta +b=0\dots ...\left(2\right)$

Since $A_n={\alpha }^n+{\beta }^n$.

$$\begin{gathered} A_{n+1}–a{A}_n+b{A}_{n-1}={\alpha }^{n+1}+{\beta }^{n+1}–a({\alpha }^n+{\beta }^n)+b({\alpha }^{n-1}+{\beta }^{n-1}) \\ \Rightarrow A_{n+1}–a{A}_n+b{A}_{n-1}={\alpha }^{n-1}({\alpha }^2–a\alpha +b)+{\beta }^{n-1}(-a\beta +b) \\ \Rightarrow A_{n+1}–a{A}_n+b{A}_{n-1}={\alpha }^{n-1}\left(0\right)+{\beta }^{n-1}\left(0\right) \\ \Rightarrow A_{n+1}–a{A}_n+b{A}_{n-1}=0 \end{gathered}$$

Hence the correct option is C.