A) \[{{V}_{n+1}}=\alpha {{V}_{n}}-b{{V}_{n-1}}\]
B) \[{{V}_{n+1}}=b{{V}_{n}}+a{{V}_{n-1}}\]
C) \[{{V}_{n+1}}=b{{V}_{n}}-a{{V}_{n-1}}\]
D) \[{{V}_{n+1}}=b{{V}_{n}}+b{{V}_{n-1}}\]
Correct Answer: A
Solution :
We have, the equation \[{{x}^{2}}-ax+b=0\] ...(i) On multiplying by\[{{x}^{n-1}},\]we get \[{{x}^{n-1}}-a{{x}^{n}}+b{{x}^{n-1}}=0\] ...(ii) Since,\[\alpha ,\beta \]are the roots of Eq. (i), therefore \[\alpha ,\beta \]are the roots of Eq. (ii), also and they will satisfy Eq. (ii). \[\therefore \] \[{{\alpha }^{n+1}}-a{{\alpha }^{n}}+b{{\alpha }^{n-1}}=0\] ...(iii) and \[{{\beta }^{n+1}}-a{{\beta }^{n}}+b{{\beta }^{n-1}}=0\] ...(iv) Adding Eqs. (iii) and (iv), \[{{\alpha }^{n+1}}+{{\beta }^{n+1}}-a(\alpha +{{\beta }^{n}})\] \[+b({{\alpha }^{n-1}}+{{\beta }^{n-1}})=0\] \[\Rightarrow \]\[{{v}_{n+1}}-a{{v}_{n}}+b{{v}_{n-1}}=0\] \[(\because {{\alpha }^{n}}+{{\beta }^{n}}={{v}_{n}})\] \[\Rightarrow \] \[{{v}_{n+1}}=a{{v}_{n}}-b{{v}_{n-1}}\]
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