RAJASTHAN ­ PET Rajasthan PET Solved Paper-2007

  • question_answer The sum of the sequence\[1+\frac{1}{4.2!}+\frac{1}{16.4!}+\frac{1}{64.6!}+......\infty \]is

    A)  \[\frac{e-1}{\sqrt{e}}\]

    B)  \[\frac{e+1}{\sqrt{e}}\]

    C)  \[\frac{e-1}{2\sqrt{e}}\]

    D)  \[\frac{e+1}{2\sqrt{e}}\]

    Correct Answer: D

    Solution :

     We know that, \[{{e}^{x}}=1+x+\frac{{{x}^{2}}}{2!}+\frac{{{x}^{3}}}{3!}+\frac{{{x}^{4}}}{4!}+....\] and     \[{{e}^{-x}}=1-x+\frac{{{x}^{2}}}{2!}-\frac{{{x}^{3}}}{3!}+\frac{{{x}^{4}}}{4!}-....\] \[\Rightarrow \] \[\frac{{{e}^{x}}+{{e}^{-x}}}{2}=1+\frac{{{x}^{2}}}{2!}+\frac{{{x}^{4}}}{4!}+\frac{{{x}^{6}}}{6!}+....\] Put \[x=\frac{1}{2}\] \[\frac{{{e}^{1/2}}+{{e}^{-1/2}}}{2}=1+{{\left( \frac{1}{2} \right)}^{2}}.\frac{1}{2!}+{{\left( \frac{1}{2} \right)}^{4}}\frac{1}{4!}+....\] \[\Rightarrow \] \[\frac{e+1}{2\sqrt{e}}=1{{\left( \frac{1}{2} \right)}^{2}}.\frac{1}{2!}+{{\left( \frac{1}{2} \right)}^{4}}\frac{1}{4!}+...\]

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