JEE Main & Advanced Mathematics Other Series Question Bank Critical Thinking

  • question_answer In  the expansion of  \[\frac{1+x}{1\,!}+\frac{{{(1+x)}^{2}}}{2\,!}+\frac{{{(1+x)}^{3}}}{3\,!}+.....,\] the coefficient of \[{{x}^{n}}\] will be

    A) \[\frac{1}{n\,!}\]

    B) \[\frac{1}{n\,!}+\frac{1}{(n+1)\,!}\]

    C) \[\frac{e}{n\,!}\]

    D) \[e\,\left[ \frac{1}{n\,!}+\frac{1}{(n+1)\,!} \right]\]

    Correct Answer: C

    Solution :

    \[\frac{1+x}{1!}+\frac{{{(1+x)}^{2}}}{2!}+\frac{{{(1+x)}^{3}}}{3!}+....\infty \] \[={{e}^{1+x}}-1=e.\,\,{{e}^{x}}-1\] \[=-1+e\left\{ 1+x+\frac{{{x}^{2}}}{2!}+\frac{{{x}^{3}}}{3!}+.... \right\}\] \ The coefficient of  \[{{x}^{n}}=e\frac{1}{n!}\].

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