JEE Main & Advanced Mathematics Binomial Theorem and Mathematical Induction Question Bank Binomial theorem for any index

  • question_answer
    Coefficient of \[{{x}^{r}}\] in the expansion of \[{{(1-2x)}^{-1/2}}\] is                                [Kurukshetra CEE 2001]

    A) \[\frac{(2r)\,!}{{{(r\,!)}^{2}}}\]

    B) \[\frac{(2r)\,!}{{{2}^{r}}{{(r!)}^{2}}}\]

    C) \[\frac{(2r)!}{{{(r!)}^{2}}{{2}^{2r}}}\]

    D) \[\frac{(2r)!}{{{2}^{r}}.(r+1)!.(r-1)!}\]

    Correct Answer: B

    Solution :

    Coefficient of \[{{x}^{r}}=\] \[\frac{\left( -\frac{1}{2} \right)\,\left( -\frac{1}{2}-1 \right)\,\,\left( -\frac{1}{2}-2 \right)...\left( -\frac{1}{2}-r+1 \right)}{r!}{{(-2)}^{r}}\] \[\frac{1\,.\,3\ .\,5...(2r-1)}{{{2}^{r}}}\,.\,\frac{{{(-1)}^{r}}{{(-1)}^{r}}{{2}^{r}}}{r!}\] = \[\frac{1\,.\,3\,.\,5...(2r-1)}{r!}=\frac{2r!}{r!\,\,r!\,\,{{2}^{r}}}\].


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