The sum of the infinite series`(2^(2))/(2!) + (2^(4))/(4!) + (2^(6))/(6!) + `... Is equal to

Options

A. `(e^(2) + 1)/(2e)`
B. `(e^(4) + 1)/(2e^(2))`
C. `((e^(2) - 1)^(2))/(2e^(2))`
D. `(e^(2) + 1)^(2))/(2e^(2))`

Answer

To find the sum of the infinite series

\[ S = \frac{2^2}{2!} + \frac{2^4}{4!} + \frac{2^6}{6!} + \ldots \]

we can recognize that this series resembles the Taylor series expansion for the exponential function \( e^x \), which is given by:

\[ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \]

However, our series only includes the even-indexed terms. To isolate these even terms, we can use the following approach:

1. **Recognize the Series**: The series can be expressed as:

\[ S = \sum_{n=1}^{\infty} \frac{(2^2)^{n}}{(2n)!} = \sum_{n=1}^{\infty} \frac{2^{2n}}{(2n)!} \]

2. **Use the Exponential Function**: We know that the series for \( e^x \) can be split into even and odd parts. The even part can be represented as:

\[ \frac{e^x + e^{-x}}{2} = \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!} \]

3. **Substituting \( x = 2 \)**: To find our series, we substitute \( x = 2 \):

\[ \frac{e^2 + e^{-2}}{2} = \sum_{n=0}^{\infty} \frac{2^{2n}}{(2n)!} \]

4. **Adjust for the Starting Index**: Our original series starts from \( n=1 \), so we need to subtract the first term (which corresponds to \( n=0 \)) from the above equation:

\[ S = \frac{e^2 + e^{-2}}{2} - 1 \]

5. **Simplifying the Expression**: Now we can simplify this expression:

\[ S = \frac{e^2 + e^{-2}}{2} - 1 \]

6. **Final Calculation**: We can express \( e^2 + e^{-2} \) in a more useful form:

\[ S = \frac{e^2 + \frac{1}{e^2}}{2} - 1 = \frac{e^2 + 1/e^2 - 2}{2} \]

This can be recognized as:

\[ S = \frac{(e - \frac{1}{e})^2}{2} \]

Thus, the final answer for the sum of the infinite series is:

\[ S = \frac{(e^2 + 1)}{2} - 1 \]