Question

$\frac{\left({\displaystyle \frac{1}{2!}}+{\displaystyle \frac{1}{4!}}+{\displaystyle \frac{1}{6!}}+...\right)}{\left(1+\frac{1}{3!}+\frac{1}{5!}+...\right)}$equals

• A. $e+1$
• B. $\frac{(e-1)}{(e+1)}$
• C. $e-1$
• D. None of these

Answer 1

The correct option is B

$\frac{(e-1)}{(e+1)}$

Explanation for the correct option:

Step 1. Put $x=1$ and $x=-x$ in $e^x$expansion:

$\Rightarrow e=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+....\infty$ ……….(1) $\left[\mathbf{\because }{\mathit{e}}^{\mathbf{x}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{}\mathbf{+}\mathbf{}\frac{\mathbf{x}}{\mathbf{1}\mathbf{!}}\mathbf{}\mathbf{+}\mathbf{}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{2}\mathbf{!}}\mathbf{}\mathbf{+}\mathbf{}\frac{{\mathbf{x}}^{\mathbf{3}}}{\mathbf{3}\mathbf{!}}\mathbf{}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\right]$

$e^{-x}=\frac{1}{0!}-\frac{x}{1!}+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}.....\infty$

Put $x=1$in above equation

$\Rightarrow e^{-1}=\frac{1}{0!}-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}.....\infty$ ……….(2)

Step 2. Add equation (1) and (2)

$\Rightarrow$ $\left(e+e^{-1}\right)=2\left[1+\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+......+\infty \right]$

$\Rightarrow$ $\left(\frac{e+e^{-1}}{2}\right)=1+\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+......+\infty$

$\Rightarrow \left(\frac{e+e^{-1}}{2}\right)-1=\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+......+\infty$ ………...(3)

Step 3. Subtract equation (2) form equation (1)

$\Rightarrow$ $\left(e-e^{-1}\right)=2\left[\frac{1}{1!}+\frac{1}{3!}+\frac{1}{5!}+.....+\infty \right]$

$\Rightarrow \left(\frac{e-e^{-1}}{2}\right)=\frac{1}{1!}+\frac{1}{3!}+\frac{1}{5!}+.....+\infty$ ………..(4)

Step 4. Divide equation (3) and (4)

$\Rightarrow \frac{\left({\displaystyle \frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+...}\right)}{\left(1+\frac{1}{3!}+\frac{1}{5!}+...\right)}=\left(\frac{e+e^{-1}-2}{e-e^{-1}}\right)$

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

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

$=\frac{(e-1)}{(e+1)}$

Hence, Option ‘B’ is Correct.