• # 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)  $\frac{e+1}{2\sqrt{e}}$

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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