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JEE Main & Advanced Mathematics Sequence & Series Question Bank Exponential series

  • question_answer
    \[1+\frac{a-bx}{1\,!}+\frac{{{(a-bx)}^{2}}}{2\,!}+\frac{{{(a-bx)}^{3}}}{3\,!}+....\infty =\]

    A) \[{{e}^{a-bx}}\]

    B) \[{{e}^{a-bx}}-1\]

    C) \[1+a{{\log }_{e}}(a-bx)\]

    D) \[{{e}^{-bx}}\]

    Correct Answer: A

    Solution :

    \[1+\frac{(a-bx)}{1\ !}+\frac{{{(a-bx)}^{2}}}{2\ !}+\frac{{{(a-bx)}^{3}}}{3\ !}+......={{e}^{a-bx}}\].


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