Question

The sum of series $\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+\dots$ is

• A. $\frac{(e^2-1)}{2}$
• B. $\frac{(e-1{)}^2}{2e}$
• C. $\frac{(e^2-1)}{2e}$
• D. $\frac{(e^2-2)}{e}$

Answer 1

The correct option is B $\frac{(e-1{)}^2}{2e}$

We know that $e=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...\infty$ ...(1)
$e^{-1}=1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-...\infty$ ...(2)
On adding equation (1) and (2)
$e+e^{-1}=2+\frac{2}{2!}+\frac{2}{4!}+...\infty$
$\Rightarrow \frac{e^2+1}{e}-2=\frac{2}{2!}+\frac{2}{4!}+...\infty$
$\Rightarrow \frac{e^2+1-2e}{e}=2[\frac{1}{2!}+\frac{1}{4!}+...\infty ]$
$\Rightarrow \frac{{(e-1)}^2}{2e}=\frac{1}{2!}+\frac{1}{4!}+...\infty$