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) \[{{V}_{n+1}}=a{{V}_{n-1}}-b{{V}_{n}}\]
Correct Answer: C
Solution :
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 \[{{V}_{2}}=a{{V}_{1}}-b{{V}_{0}}={{a}^{2}}-2b\]
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