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

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

    A) \[x=\]

    B) \[{{e}^{a}}\]

    C) \[\frac{e}{{{e}^{b/a}}}\]

    D) \[\frac{e}{{{e}^{a/b}}}\]

    Correct Answer: C

    Solution :

    \[1+\frac{a-b}{a}+\frac{1}{2\ !}{{\left( \frac{a-b}{a} \right)}^{2}}\ +\frac{1}{3\ !}{{\left( \frac{a-b}{a} \right)}^{3}}+........\] \[=1+\frac{\left( \frac{a-b}{a} \right)}{1\ !}+\frac{{{\left( \frac{a-b}{a} \right)}^{2}}}{2\ !}+\frac{{{\left( \frac{a-b}{a} \right)}^{3}}}{3\ !}+.......\]       \[={{e}^{(a-b)/a}}={{e}^{1-(b/a)}}={{e}^{1}}.{{e}^{-b/a}}=\frac{e}{{{e}^{b/a}}}\] .


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