Question

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

Answer: (B)

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