Question

$\frac{\left(1+{\displaystyle \frac{1}{2!}}+{\displaystyle \frac{2}{3!}}+{\displaystyle \frac{2^2}{4!}}+{\displaystyle \frac{2^3}{5!}}+\dots \right)}{\left(1+\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+...\right)}$ is equal to

• A. $\frac{e}{4}$
• B. $8e$
• C. $\frac{e}{2}$
• D. None of these

Answer 1

The correct option is C

$\frac{e}{2}$

Explanation for the correct option:

Step 1. Find the sum of given series:

$\frac{\left(1+{\displaystyle \frac{1}{2!}}+{\displaystyle \frac{2}{3!}}+{\displaystyle \frac{2^2}{4!}}+{\displaystyle \frac{2^3}{5!}}+\dots \right)}{\left(1+\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+...\right)}$ ….…(i)

As we know, ${\mathit{e}}^{\mathbf{x}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{+}\frac{\mathbf{x}}{\mathbf{1}\mathbf{!}}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{2}\mathbf{!}}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{3}}}{\mathbf{3}\mathbf{!}}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{4}}}{\mathbf{4}\mathbf{!}}\mathbf{+}\mathbf{\dots }$

Step 2. Put $x=2$ in $e^x$ expression:

$\Rightarrow e^2=1+\frac{2}{1!}+\frac{2^2}{2!}+\frac{2^3}{3!}+\frac{2^4}{4!}+\dots$ ………(ii)

Step 3. Multiply and divide numerator of equation (i) by $2^2$:

$\frac{\left(1+{\displaystyle \frac{1}{2!}}+{\displaystyle \frac{2}{3!}}+{\displaystyle \frac{2^2}{4!}}+{\displaystyle \frac{2^3}{5!}}+\dots \right)}{\left(1+\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+...\right)}$$=\left(\frac{1}{2^2}\right)\left[\frac{\left(2^2+{\displaystyle \frac{2^2}{2!}}+{\displaystyle \frac{2^3}{3!}}+{\displaystyle \frac{2^4}{4!}}+{\displaystyle \frac{2^5}{5!}}+\dots \right)}{\left(1+\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+\dots \right)}\right]$

$=\left(\frac{1}{2^2}\right)\left[\frac{(2^2+e^2-3)}{{\displaystyle \frac{(e+e^{-1})}{2}}}\right]$ $\left[\mathbf{\because }\mathit{e}\mathbf{+}{\mathit{e}}^{\mathbf{-}\mathbf{1}}\mathbf{=}\mathbf{2}\mathbf{(}\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}\mathbf{!}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{4}\mathbf{!}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{6}\mathbf{!}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{)}\right]$

$=\left(\frac{1}{2^2}\right)\left[\frac{2(1+e^2)}{\left({\displaystyle \frac{(e^2+1)}{e}}\right)}\right]$

$=\frac{\mathbf{e}}{\mathbf{2}}$

Hence, Option ‘C’ is Correct.