Question

If $\alpha$ and $\beta$ are the roots of $x^2-ax+b=0$ and if ${\alpha }^n+{\beta }^n=V_n$, then

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

Answer 1

Answer: (C)

$V_{n+1}=a{V}_n-b{V}_{n-1}$

Multiplying $x^2-ax+b=0$ by $x^{n-1}$, we get
$x^{n+1}-a{x}^n+b{x}^{n-1}=0....\left(i\right)$
$\alpha ,\beta$ are roots of $x^2-ax+b=0$, therefore
they will satisfy (i)
Also, ${\alpha }^{n+1}-a{\alpha }^{n-1}=0....\left(ii\right)$
and ${\beta }^{n+1}-a{\beta }^n+b{\beta }^{n-1}=0....\left(ii\right)$
On adding eqs. (ii) and (iii), we get
$({\alpha }^{n+1}+{\beta }^{n+1})-a({\alpha }^n+{\beta }^n)+b({\alpha }^{n-1}+{\beta }^{(n-1)})=0$
$\Rightarrow V_{n+1}-a{V}_n+b{V}_{n-1}=0$
$\Rightarrow V_{n+1}=a{V}_n-b{V}_{n-1}$