Q. The sum of the infinite series $\frac{2^{2}}{2!} + \frac{2^{4}}{4!} + \frac{2^{6}}{6!} + \ldots$ is equal to
A) $\frac{e^{2} + 1}{2e}$
B) $\frac{e^{4} + 1}{2e^{2}}$
C) $\frac{\left( e^{2} - 1 \right)^{2}}{2e^{2}}$
D) $\frac{\left( e^{2} + 1 \right)^{2}}{2e^{2}}$
Explanation
(c)
Now, $e^{2} = 1 + \frac{2}{1!} + \frac{2^{2}}{2!} + \frac{2^{3}}{3!} + ...$
and $e^{- 2} = 1 - \frac{2}{1!} + \frac{2^{2}}{2!} - \frac{2^{3}}{3!} + ...$
$\Rightarrow e^{2} + e^{- 2} = 2\left\lbrack 1 + \frac{2^{2}}{2!} + \frac{2^{4}}{4!} + ... \right\rbrack$
$\Rightarrow \frac{e^{2} + e^{- 2}}{2} - 1 = \left\lbrack \frac{2^{2}}{2!} + \frac{2^{4}}{4!} + ... \right\rbrack$
$\Rightarrow \frac{\begin{array}{r}
\ \\
e^{4} + 1 - 2e^{2}
\end{array}}{2e^{2}} = \left\lbrack \frac{2^{2}}{2!} + \frac{2^{4}}{4!} + ... \right\rbrack$
$\Rightarrow \ \frac{\left( e^{2} - 1 \right)^{2}}{2e^{2}} = \frac{2^{2}}{2!} + \frac{2^{4}}{4!} + ...$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.