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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 13

  • question_answer
    If the constant term in the binomial expansion of  \[{{\left( {{x}^{2}}-\frac{1}{x} \right)}^{n}},\] \[n\in N\] is 15, then the value of n is equal to

    A)  \[4\]                        

    B)         \[6\]          

    C)  \[7\]                      

    D)         \[9\]

    Correct Answer: B

    Solution :

           [b] We have \[{{\left( {{x}^{2}}-\frac{1}{x} \right)}^{n}}\] \[{{T}_{r+1}}{{=}^{n}}{{C}_{r}}{{({{x}^{2}})}^{n-r}}{{(-1)}^{r}}{{x}^{-r}}{{=}^{n}}{{C}_{r}}\,{{x}^{2n-3r}}{{(-1)}^{r}}\] Constant term \[{{=}^{n}}{{C}_{r}}{{(-1)}^{r}}\] if \[2n=3r\] \[\Rightarrow \,\,{{\,}^{n}}{{C}_{2n/3}}{{(-1)}^{2n/3}}=15{{=}^{6}}{{C}_{4}}\] \[\Rightarrow \,\,\,n=6\]      


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