Question

The sum of the infinite series$\frac{2^2}{2!}+\frac{2^4}{4!}+\frac{2^6}{6!}+\dots$ is equal to

• A. $\frac{(e^2+1)}{2e}$
• B. $\frac{(e^4+1)}{2e^2}$
• C. $\frac{{(e^2-1)}^2}{2e^2}$
• D. None of these

Answer 1

The correct option is C

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

Explanation for the correct option:

Given: $\frac{2^2}{2!}+\frac{2^4}{4!}+\frac{2^6}{6!}+\dots$

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

Put, $x=2$

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

Put, $x=-2$

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

Adding $\left(1\right)$ and $\left(2\right)$

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

Hence, the correct option is C.