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

  • question_answer
    The value of \[\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}\sin (rx)}\] is equal to

    A) \[{{2}^{n}}\cdot {{\cos }^{n}}\frac{x}{2}\cdot \sin \frac{nx}{2}\]

    B) \[{{2}^{n}}\cdot si{{n}^{n}}\frac{x}{2}\cdot \cos \frac{nx}{2}\]

    C) \[{{2}^{n+1}}\cdot {{\cos }^{n}}\frac{x}{2}\cdot \sin \frac{nx}{2}\]

    D) \[{{2}^{n+1}}\cdot si{{n}^{n}}\frac{x}{2}\cdot \cos \frac{nx}{2}\]

    Correct Answer: A

    Solution :

    [a] \[\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}\sin \,rx=\operatorname{Im}\left( \sum\limits_{r=0}^{n}{^{n}{{C}_{r}}{{e}^{irx}}} \right)}\]
    \[=\operatorname{Im}\left( \sum\limits_{r=0}^{n}{^{n}{{C}_{r}}{{\left( {{e}^{ix}} \right)}^{r}}} \right)=\operatorname{Im}\left( {{\left( 1+{{e}^{ix}} \right)}^{n}} \right)\]
    \[=\operatorname{Im}{{(1+\cos \,x+i\,\sin \,x)}^{n}}\]
    \[=\operatorname{Im}\,{{(2\,co{{s}^{2}}\frac{x}{2}+2i\,sin\frac{x}{2}.cos\frac{x}{2})}^{n}}\]
    \[=\operatorname{Im}\,{{\left( 2\cos \frac{x}{2}\left( \cos \frac{x}{2}+i\,\,\sin \frac{x}{2} \right) \right)}^{n}}\]
    \[={{2}^{n}}.{{\cos }^{n}}\frac{x}{2}.\sin \frac{nx}{2}\]


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