Question

$\alpha ,\beta$ 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. $V_{n+1}=a{V}_n+b{V}_{n-1}$
• C. $V_{n+1}=a{V}_{n-1}-b{V}_n$
• D. $V_{n+1}=a{V}_n+a{V}_{n-1}$

Answer 1

The correct option is A

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

Multiplying $x^2-ax+b=0by{x}^{n-1}$

$x^{n+1}-a{x}^n+b{x}^{n-1}=0$ .........(i)

$\alpha ,\beta$ are roots of $x^2–ax+b=0$, therefore they will satisfy (i)
${\alpha }^{n+1}-a{\alpha }^n+b{\alpha }^{n-1}=0$ ......(ii)

and ${\beta }^{n+1}-a{\beta }^n+b{\beta }^{n-1}=0$ .......(iii)

Adding (ii) and (iii)

$({\alpha }^{n+1}+{\beta }^{n+1})-a({\alpha }^n+{\beta }^n)+b({\alpha }^{n-1}+{\beta }^{n-1})=0or{V}_{n+1}-a{V}_n+b{V}_{n-1}=0or{V}_{n+1}-a{V}_n+b{V}_{n-1}=0(Given{\alpha }^n+{\beta }^n=V_n)$

Alternate solution:

Trick: Put n = 0, 1, 2

$V_0={\alpha }^0+{\beta }^0-2,V_1=\alpha +\beta =a,{\alpha }^2+{\beta }^2=V_2=a^2-2b$
Now the option (c) $\Rightarrow V_2=a{V}_1-b{V}_0=a^2-2b$