• # question_answer 23) If $\alpha ,\beta$are the roots of ${{x}^{2}}-ax+b=0$ and if ${{\alpha }^{n}}+{{\beta }^{n}}={{V}_{n}}$, then      [RPET 1995; Karnataka CET 2000; Pb. CET 2002] 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) ${{V}_{n+1}}=a{{V}_{n-1}}-b{{V}_{n}}$

Multiplying ${{x}^{2}}-ax+b=0$by ${{x}^{n-1}}$ ${{x}^{n+1}}-a{{x}^{n}}+b{{x}^{n-1}}=0$ .....(i) $\alpha ,\beta$are roots of ${{x}^{2}}-ax+b=0$, therefore they will satisfy (i) 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) $({{\alpha }^{n+1}}+{{\beta }^{n+1}})-a({{\alpha }^{n}}+{{\beta }^{n}})+b({{\alpha }^{n-1}}+{{\beta }^{n-1}})=0$ or    ${{V}_{n+1}}-a{{V}_{n}}+b{{V}_{n-1}}=0$ or   ${{V}_{n+1}}=a{{V}_{n}}-b{{V}_{n-1}}=0$(Given ${{\alpha }^{n}}+{{\beta }^{n}}={{V}_{n}}$) Trick: Put$n=0$, $1,\,\,2$ ${{V}_{0}}={{\alpha }^{0}}+{{\beta }^{0}}=2$, ${{V}_{1}}=\alpha +\beta =a$,  ${{\alpha }^{2}}+{{\beta }^{2}}={{V}_{2}}={{a}^{2}}-2b$ Now the option C Þ ${{V}_{2}}=a{{V}_{1}}-b{{V}_{0}}={{a}^{2}}-2b$