Question

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

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

Answer 1

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

Given,

to find out the sum of series,

$\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+\dots \dots .$

as we know that,

$e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\dots \dots$

$e^{-x}=1-\frac{x}{1!}+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}-\frac{x^5}{5!}+\dots \dots$

let us assume $x$ as ${}^{\prime }{1}^{\prime }$ in both $e^x$ & $e^{-x}$ equations we get,

$e^1=1+\frac{1}{1!}+\frac{2}{2!}+\frac{3}{3!}+\frac{4}{4!}+\dots \dots$

$\Rightarrow e=1+1+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\dots ..$

$\Rightarrow e=2+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\dots ..\left(3\right)$

Similarly

$e^{-1}=1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\dots ..$

$\Rightarrow \frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\dots ..\left(4\right)$

Now, let us add eq $\left(3\right)$ & eq $\left(4\right)$ we get,

$e+\frac{1}{e}=[2+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}\dots .]+[\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\dots ..]$

$\Rightarrow [2+\frac{2}{2!}+\frac{2}{4!}+\frac{2}{6!}+\dots .]$

$\Rightarrow e+\frac{1}{e}=2[1+\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+\dots ..]$

$\Rightarrow 1+\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+\dots ..=\frac{1}{2}[e+\frac{1}{e}]$

$\Rightarrow \frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+\dots ..=\frac{1}{2}[e+\frac{1}{e}]-1$

Now, R.H.S= $\frac{1}{2}[e+\frac{1}{e}]-1$

$\Rightarrow \frac{1}{2}\left[\frac{e^2+1}{e}\right]-1\Rightarrow \frac{e^2+1}{2e}-1$

$[\because a^2-2ab+b^2=(a-b{)}^2]$

$\Rightarrow \frac{e^2+1-2e}{2e}\Rightarrow \frac{e^2-2.e.1+(1{)}^2}{2e}\Rightarrow \frac{(e-1{)}^2}{2e}$

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