Question

$If\alpha ,\beta aretherootsof{x}^2-ax+b=0andif{\alpha }^n+{\beta }^n=V_n,then$

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

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

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

$\alpha ,\beta$ are roots of $x^2$ – ax – ax + b = 0, therefore they will satisfy

(i) also ${\alpha }^{n+1}-a{\alpha }^n+b{\alpha }^{n-1}=0$ ......(ii)

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

Adding (ii) and (iii)

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

Trick: Put n = 0, 1, 2

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