JEE Main & Advanced Mathematics Other Series Question Bank Critical Thinking

  • question_answer If  \[n\] is even, then in the expansion of  \[{{\left( 1+\frac{{{x}^{2}}}{2\,!}+\frac{{{x}^{4}}}{4\,!}+...... \right)}^{2}}\], the coefficient of \[{{x}^{n}}\] is

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

    B) \[\frac{{{2}^{n}}-2}{n\,\,!}\]

    C) \[\frac{{{2}^{n-1}}-1}{n\,!}\]

    D) \[\frac{{{2}^{n-1}}}{n\,!}\]

    Correct Answer: D

    Solution :

    \[{{\left( 1+\frac{{{x}^{2}}}{2!}+\frac{{{x}^{4}}}{4!}+.... \right)}^{2}}={{\left( \frac{{{e}^{x}}+{{e}^{-x}}}{2} \right)}^{2}}\]                                  \[=\frac{1}{4}({{e}^{2x}}+{{e}^{-2x}}+2)\] \[=\frac{1}{4}\left\{ 2\,\left( 1+\frac{{{(2x)}^{2}}}{2!}+\frac{{{(2x)}^{4}}}{4!}+.... \right)+2 \right\}\] \ The coefficient of  \[{{x}^{n}}\](n even)\[=\frac{1}{2}\left\{ \frac{{{2}^{n}}}{n!} \right\}=\frac{{{2}^{n-1}}}{n!}\].

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