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JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Afternoon)

  • question_answer
    The smallest natural number n, such that the coefficient of x in the expansion of\[{{\left( {{x}^{2}}+\frac{1}{{{x}^{3}}} \right)}^{n}}\]is \[^{n}{{C}_{23}},\]is : [JEE Main 10-4-2019 Afternoon]

    A) 35                               

    B) 38

    C) 23                               

    D) 58

    Correct Answer: B

    Solution :

    \[{{T}_{r}}=\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}}{{x}^{2n-2r}}.{{x}^{-3r}}\] \[2n-5r=1\]\[\Rightarrow 2n=5r+1\] for \[r=15.\]\[n=38\] smallest value of n is 38.


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