JEE Main & Advanced Mathematics Binomial Theorem and Mathematical Induction Question Bank Mock Test - Binomial Theorem

  • question_answer
    If \[{{x}^{m}}\]occurs in the expansion of \[{{(x+1/{{x}^{2}})}^{2n}}\], then the coefficient of \[{{x}^{m}}\]is

    A) \[\frac{(2n)!}{(m)!(2n-m)!}\]      

    B) \[\frac{(2n)!3!3!}{(2n-m)!}\]

    C) \[\frac{(2n)!}{\left( \frac{2n-m}{3} \right)!\left( \frac{4n+m}{3} \right)!}\]

    D) none of these

    Correct Answer: C

    Solution :

    [c]
    \[{{T}_{r+1}}{{=}^{2n}}{{C}_{r}}{{x}^{2n-r}}{{\left( \frac{1}{{{x}^{2}}} \right)}^{r}}{{=}^{2n}}{{C}_{r}}{{x}^{2n-3r}}\]
    \[This\,\,contains\text{ }{{x}^{m}}\,.If\text{ }2n-3r=m,\,\,then\]
    \[r=\frac{2n-m}{3}\]
    \[\Rightarrow Coefficient\,\,of\text{ }{{x}^{m}}{{=}^{2n}}{{C}_{r}},\]
    \[r=\frac{2n-m}{3}\]
    \[=\frac{2n!}{(2n-r)!r!}=\frac{2n!}{\left( 2n-\frac{2n-m}{3} \right)!\left( \frac{2n-m}{3} \right)!}\]
    \[=\frac{2n!}{\left( \frac{4n+m}{3} \right)!\left( \frac{2n-m}{3} \right)!}\]
     


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