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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
    The coefficient of \[{{x}^{39}}\] in the expansion of \[{{\left( {{x}^{4}}-\frac{1}{{{x}^{3}}} \right)}^{15}}\] is   [MP PET 2001]

    A) - 455

    B) - 105

    C) 105

    D) 455

    Correct Answer: A

    Solution :

    \[{{T}_{r+1}}=\,{}^{15}{{C}_{r}}{{({{x}^{4}})}^{15-r}}{{(-1/{{x}^{3}})}^{r}}\]= \[{{(-1)}^{r}}\,\,{}^{15}{{C}_{r}}{{(x)}^{60-7r}}\] For coefficient of\[{{x}^{39}}\],  \[60-7r=39\Rightarrow r=3\] \ \[{{T}_{4}}={}^{15}{{C}_{3}}{{({{x}^{4}})}^{12}}{{(-1/{{x}^{3}})}^{3}}\]= \[-455\,{{x}^{39}}\] Hence the required coefficient is - 455.


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