JEE Main & Advanced Mathematics Binomial Theorem and Mathematical Induction Question Bank General term, Coefficient of any power of x, Independent term, Middle term and Greatest term and Greatest coefficient

  • question_answer
    \[{{16}^{th}}\] term in the expansion of \[{{(\sqrt{x}-\sqrt{y})}^{17}}\] is

    A) \[136x{{y}^{7}}\]

    B) \[136xy\]

    C) \[-136x{{y}^{15/2}}\]

    D) \[-136x{{y}^{2}}\]

    Correct Answer: C

    Solution :

    \[{{T}_{16}}={{\,}^{17}}{{C}_{15}}{{(\sqrt{x})}^{2}}{{(-\sqrt{y})}^{15}}\] \[=-\frac{17\times 16}{2\times 1}\times x{{y}^{15/2}}=-136x{{y}^{15/2}}\]


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