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JEE Main & Advanced AIEEE Solved Paper-2002

  • question_answer
    The     coefficient    of    \[{{x}^{5}}\]  in \[{{(1+2x+3{{x}^{2}}+....)}^{-3/2}}\] is   AIEEE  Solved  Paper-2002

    A)                                         21                              

    B)           25                              

    C)           26                              

    D)           None of these

    Correct Answer: D

    Solution :

    Since, \[{{(1+2x+3{{x}^{2}}+....)}^{-3/2}}={{[{{(1-x)}^{-2}}]}^{-3/2}}\]                                                    \[={{(1-x)}^{3}}\] So, coefficient of \[{{x}^{5}}\] in \[{{(1+2x+3{{x}^{2}}+...)}^{-3/2}}\]                    = Coefficient of \[{{x}^{n}}\] in \[{{(1-x)}^{3}}=0\]


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