J & K CET Engineering J and K - CET Engineering Solved Paper-2006

  • question_answer
    The coefficient of \[{{x}^{-17}}\]in the expansion of \[{{\left( {{x}^{4}}-\frac{1}{{{x}^{3}}} \right)}^{15}}\]is

    A)  \[^{15}{{C}_{11}}\]

    B)  \[^{15}{{C}_{12}}\]

    C)  \[^{-15}{{C}_{11}}\]

    D)  \[^{-15}{{C}_{3}}\]

    Correct Answer: C

    Solution :

    General term, \[{{T}_{r+1}}={{(-1)}^{r\,\,15}}{{C}_{r}}\,{{({{x}^{4}})}^{15-r}}.{{\left( \frac{1}{{{x}^{3}}} \right)}^{r}}\] \[={{(-1)}^{r}}\,{{\,}^{15}}{{C}_{r}}.\,{{x}^{60-7r}}\] For the coefficient of \[{{x}^{-17}},\] put \[60-7r=-17\] \[\Rightarrow \] \[60+17=7r\] \[\Rightarrow \] \[7r=77\] \[\Rightarrow \] \[r=11\] Now, coefficient of \[{{x}^{-17}}\] \[={{(-1)}^{11}}{{\,}^{15}}{{C}_{11}}={{-}^{15}}{{C}_{11}}\]


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