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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2013

  • question_answer
        If\[\alpha \]and\[\beta \]be the roots of the equation\[{{x}^{2}}-ax+b=0\]and\[{{A}_{n}}={{\alpha }^{n}}+{{\beta }^{n}}\]. Then, \[{{A}_{n+1}}-a{{A}_{n}}+b{{A}_{n-1}}\]is equal to

    A)  \[-a\]                                   

    B)  b

    C)  0                                            

    D)  \[a-b\]

    Correct Answer: C

    Solution :

                     We have,                                                                                                         \[\alpha +\beta =a\] and\[\alpha \beta =0.\] Now, \[a{{A}_{n}}-n{{A}_{n-1}}=(\alpha +\beta )({{\alpha }^{n}}+{{\beta }^{n}})\]                                                 \[-\alpha \beta ({{\alpha }^{n-1}}+{{\beta }^{n-1}})\] \[={{\alpha }^{n+1}}+{{\alpha }^{n}}\beta +\alpha {{\beta }^{n}}+{{\beta }^{n+1}}-{{\alpha }^{n}}\beta -\alpha {{\beta }^{n}}\] \[={{\alpha }^{n+1}}+{{\beta }^{n+1}}\] \[={{A}_{n+1}}\]


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