Question

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

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

Answer 1

Answer: (B)

$\frac{{\left(e-1\right)}^2}{2e}$

Solution -

$\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+----$

$e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+----$

$e=1+\frac{1}{1!}+\frac{1}{2!}+---$

$e^{-1}=1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+---$

$e+e^{-1}=2(1+\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+---)$

$\frac{e^2+1}{2e}=1+\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+-----$

$\frac{(e-1{)}^2}{2e}=\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+-----$

B is correct