Question

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

• A. $1+\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+....\infty$
• B. 0
• C. 1
• D. none of the above

Answer 1

Answer: (A)

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