Question

The sum of the infinite series $\dfrac{{{2}^{2}}}{2!}+\dfrac{{{2}^{4}}}{4!}+\dfrac{{{2}^{6}}}{6!}+...$ is equal to
A. $\dfrac{{{e}^{2}}+1}{2e}$
B. $\dfrac{{{e}^{4}}+1}{2{{e}^{2}}}$
C. $\dfrac{{{({{e}^{2}}-1)}^{2}}}{2{{e}^{2}}}$
D. $\dfrac{{{({{e}^{2}}+1)}^{2}}}{2{{e}^{2}}}$

Answer 1

Hint: In this question, we are to find the sum of the given infinite series. Here the series denotes an exponential series. So, by applying the exponential theorem, we can find the required sum to infinite series. To get the required series, we need to substitute $x=2$ in the theorem, so that we can frame the required series. Then, on simplifying the exponential terms, we get the required sum.

Formula Used: Exponential theorem:
The exponential series is written as
${{e}^{x}}=1+\dfrac{x}{1!}+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{3}}}{3!}+...$
We can write it as
${{e}^{x}}=\sum\limits_{n=0}^{\infty }{\dfrac{{{x}^{n}}}{n!}}$
And for the negative power, we have
${{e}^{-x}}=1-\dfrac{x}{1!}+\dfrac{{{x}^{2}}}{2!}-\dfrac{{{x}^{3}}}{3!}+...=\sum\limits_{n=0}^{\infty }{{{\left( -1 \right)}^{n}}\dfrac{{{x}^{n}}}{n!}}$
With these expansions, we can find
$\sinh x=\dfrac{{{e}^{x}}-{{e}^{-x}}}{2}=x+\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{5}}}{5!}+....$
And
$\cosh x=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}=1+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{4}}}{4!}+....$

Complete step by step solution: Given infinite series is
$\dfrac{{{2}^{2}}}{2!}+\dfrac{{{2}^{4}}}{4!}+\dfrac{{{2}^{6}}}{6!}+...\text{ }(1)$
But we have the exponential theorem as
$\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}=1+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{4}}}{4!}+....\text{ }(2)$
On simplifying the expansion at (2), we get
$\begin{align}
  & \dfrac{{{e}^{x}}+{{e}^{-x}}}{2}=1+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{4}}}{4!}+.... \\
 & \Rightarrow \dfrac{{{e}^{x}}+{{e}^{-x}}}{2}-1=\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{4}}}{4!}+.... \\
\end{align}$
Now, by substituting $x=2$ in the above expansion, we get
$\begin{align}
  & \dfrac{{{e}^{x}}+{{e}^{-x}}}{2}-1=\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{4}}}{4!}+... \\
 & \Rightarrow \dfrac{{{e}^{2}}+{{e}^{-2}}}{2}-1=\dfrac{{{2}^{2}}}{2!}+\dfrac{{{2}^{4}}}{4!}+...\text{ }(3) \\
\end{align}$
On comparing (1) and (3), we can write the sum to the given infinite series as
\[\begin{align}
  & {{S}_{\infty }}=\dfrac{{{e}^{2}}+{{e}^{-2}}}{2}-1 \\
 & \text{ }=\dfrac{{{e}^{2}}+\dfrac{1}{{{e}^{2}}}}{2}-1 \\
 & \text{ }=\dfrac{{{\left( {{e}^{2}} \right)}^{2}}+1}{2{{e}^{2}}}-1 \\
 & \text{ }=\dfrac{{{\left( {{e}^{2}} \right)}^{2}}-2{{e}^{2}}+1}{2{{e}^{2}}} \\
 & \text{ }=\dfrac{{{\left( {{e}^{2}}-1 \right)}^{2}}}{2{{e}^{2}}} \\
\end{align}\]

Option ‘C’ is correct

Note: Here, by observing the given infinite series we come to know that it is an exponential series. So, by applying the exponential theorem, we get the required sum to the infinite series. We can also calculate the sum to infinity by applying other sequence and series methods. But their evaluation is a little complicated. So, it is good to remember the direct formula, i.e., the exponential theorem to solve this kind of series.