JEE Main & Advanced Mathematics Other Series Question Bank Critical Thinking

  • question_answer 9) In  the expansion of  \[\frac{a+bx+c{{x}^{2}}}{{{e}^{x}}}\], the coefficient of \[{{x}^{n}}\] will be

    A) \[\frac{a\,{{(-1)}^{n}}}{n\,!}+\frac{b{{(-1)}^{n-1}}}{(n-1)\,!}+\frac{c{{(-1)}^{n-2}}}{(n-2)\,!}\]

    B) \[\frac{a}{n\,!}+\frac{b}{(n-1)\,!}+\frac{c}{(n-2)\,!}\]

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

    D) None of these

    Correct Answer: A

    Solution :

    \[(a+bx+c{{x}^{2}}){{e}^{-x}}\]\[=(a+bx+c{{x}^{2}})\left\{ 1-\frac{x}{1!}+\frac{{{x}^{2}}}{2!}-\frac{{{x}^{3}}}{3!}+.... \right\}\] Thus coefficient of xn =\[a\,.\,\frac{{{(-1)}^{n}}}{n\,!}+b\frac{{{(-1)}^{n-1}}}{(n-1)\,!}+c\frac{{{(-1)}^{n-2}}}{(n-2)\,!}\].

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