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Q. $ \frac{{{e}^{2}}+1}{2e} $ is equal to:

Bihar CECEBihar CECE 2002

Solution:

$\frac{e^{2}+1}{2 e}=\frac{e+e^{-1}}{2}$
$=\frac{1}{2}\left[\left(1+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}+\ldots\right)+\left(1-\frac{1}{1 !}+\frac{1}{2 !}-\frac{1}{3 !}+\ldots\right)\right]$
$=\frac{1}{2}\left[2\left(1+\frac{1}{2 !}+\frac{1}{4 !}+\ldots\right)\right]$
$=1+\frac{1}{2 !}+\frac{1}{4 !}+\ldots$