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KVPY Sample Paper KVPY Stream-SX Model Paper-3

  • question_answer
    Let \[\alpha \] and \[\beta \] be the roots of \[{{x}^{2}}-x-1=0,\] with \[\alpha \,>\,\beta .\] For all positive integer n, define \[{{a}_{n}}\,=\,\frac{{{\alpha }^{n}}\,-\,{{\beta }^{n}}}{\alpha \,-\,\beta },\,n\ge \,1\], \[{{b}_{1}}\,=\,1\] and \[{{b}_{n}}\,=\,{{a}_{n-1}}\,+\,{{a}_{n}}+{{1}^{,\,}}\,n\,\,\le \,2\]. Then which of the following options is/are correct?

    A) \[{{a}_{1}}+{{a}_{2}}+{{a}_{3}}+{{a}_{n}}={{a}_{n+2}}-\,1\] for all \[n\,\,\ge \,\,1\]

    B) \[{{b}_{n}}={{a}^{n}}+{{\beta }^{n}}\] for all \[n\,\ge \,1\]

    C) \[\sum\limits_{n\,=\,1}^{\infty }{\frac{{{b}_{n}}}{{{10}^{n}}}\,=\,\frac{8}{89}}\]

    D) \[\sum\limits_{n\,=\,1}^{\infty }{\frac{{{a}_{n}}}{{{10}^{n}}}\,=\,\frac{10}{89}}\]

    Correct Answer: A , B , D

    Solution :

    \[{{x}^{2}}-x-1=0\]
    \[{{a}_{n}}=\frac{{{\alpha }^{n}}-{{\beta }^{n}}}{\alpha -\beta }\]
    [b]        \[{{b}_{1}}=1\]
                \[{{b}_{n}}=\,{{a}_{n-1}}+{{a}_{n+1}}\]
                \[\alpha =\frac{1+\sqrt{5}}{2},\beta =\frac{1-\sqrt{5}}{2}\]
                \[{{b}_{n}}=\frac{{{\alpha }^{n\,-\,1}}-{{\beta }^{n\,-\,1}}}{\alpha -\beta }+\frac{{{\alpha }^{n+1}}+{{\beta }^{n+1}}}{\alpha -\beta }\]\[=\frac{{{\alpha }^{n-1}}(1+{{\alpha }^{2}})-{{\beta }^{n-1}}(1+{{\beta }^{2}})}{\alpha -\beta }\]\[=\,\frac{{{\alpha }^{n\,-\,1}}\left( \frac{5+\sqrt{5}}{2} \right)-{{\beta }^{n\,-\,1}}\left( \frac{5-\sqrt{5}}{2} \right)}{\alpha -\beta }\]
                \[=\,\frac{\sqrt{5}{{\alpha }^{n}}+\sqrt{5}{{\beta }^{n}}}{\alpha -\beta }={{\alpha }^{n}}+{{\beta }^{n}}\]
    (i)         \[{{a}_{1}}+{{a}_{2}}+{{a}_{3}}+...+{{a}_{n}}\]\[=\,\frac{(\alpha +{{\alpha }^{2}}+...+{{\alpha }^{n}})-(\beta +{{\beta }^{2}}+...{{\beta }^{n}})}{\alpha -\beta }\]\[=\,\frac{\frac{\alpha (1-{{\alpha }^{n}})}{1-\alpha }-\frac{\beta (1-{{\beta }^{n}})}{1-\beta }}{\alpha -\beta }\]
                \[{{\alpha }^{2}}-\alpha -1=0\]
                \[{{\alpha }^{2}}-1=\alpha \]
                \[\alpha +1=\frac{\alpha }{\alpha -1}\]\[=\,\frac{-{{\alpha }^{2}}(1-{{\alpha }^{n}})+{{\beta }^{2}}(1-{{\beta }^{n}})}{\alpha -\beta }\]\[=\frac{-{{\alpha }^{2}}+{{\alpha }^{n+2}}+{{\beta }^{2}}-{{\beta }^{n+2}}}{(\alpha -\beta )}\]\[=\,\frac{{{\alpha }^{n+2}}-{{\beta }^{n+2}}}{(\alpha -\beta )}-\,(\alpha +\beta )\]\[={{\alpha }_{n+2}}-1.\]
    [c]         \[\sum{\frac{{{b}_{n}}}{{{10}^{n}}}=\sum{\left( \frac{{{\alpha }^{n}}}{{{10}^{n}}}+\frac{{{\beta }^{n}}}{{{10}^{n}}} \right)}}\]         \[=\,\left( \frac{\alpha }{10}+\frac{{{\alpha }^{2}}}{{{10}^{2}}}+... \right)\]\[=\,\frac{\frac{\alpha }{10}}{1\,-\,\frac{\alpha }{10}}+\frac{\beta }{1\,-\,\frac{\beta }{10}}\]\[=\,\frac{\alpha }{10-\alpha }+\frac{\beta }{10-\beta }\]\[=\,\frac{10(\alpha +\beta )-2\alpha \beta }{100-10(\alpha +\beta )+\alpha \beta }\]
                \[=\,\frac{10+2}{100-10-1}=\,\frac{12}{89}\]
    [d]        \[\sum{\frac{{{a}^{n}}}{{{10}^{n}}}=\frac{1}{\alpha -\beta }\left\{ \frac{\alpha }{10-\alpha }-\frac{\beta }{10-\beta } \right\}}\]\[=\,\frac{1}{\alpha -\beta }\left\{ \frac{10(\alpha -\beta )}{89} \right\}\,=\,\frac{10}{89}.\]

    Solution :

    Same as above

    Solution :

    Same as above


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