Question

If $\alpha$ and $\beta$ are the roots of the equations $x^2-ax+b=0$ and $A_n={\alpha }^n+{\beta }^n$, then which of the following is true?

• A. $A_{n+1}=a{A}_n+b{A}_{n-1}$
• B. $A_{n+1}=b{A}_n+a{A}_{n-1}$
• C. $A_{n+1}=a{A}_n-b{A}_{n-1}$
• D. $A_{n+1}=b{A}_n-b{A}_{n-1}$

Answer 1

The correct option is C $A_{n+1}=a{A}_n-b{A}_{n-1}$

$\alpha$ and $\beta$ are roots of the equation $x^2-ax+b=0$
So, $\alpha +\beta =a$ and $\alpha \beta =b$$A_n={\alpha }^n+{\beta }^n$
$A_0={\alpha }^0+{\beta }^0=1+1=2$
$A_1={\alpha }^1+{\beta }^1=\alpha +\beta =a$
$A_2={\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta =a^2-2b=a{A}_1-b{A}_0$
$A_3={\alpha }^3+{\beta }^3=(\alpha +\beta )({\alpha }^2+{\beta }^2-\alpha \beta )=(\alpha +\beta )({\alpha }^2+{\beta }^2)-(\alpha +\beta )\left(\alpha \beta \right)=a{A}_2-b{A}_1$
Similarly,
$A_4=a{A}_3-b{A}_2$
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.
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$A_{n+1}=a{A}_n-b{A}_{n-1}$