https://file.aiquickdraw.com/mathbot/file/17298022827053k6oxchi.jpg

The series given in the image is:

\[
\sum_{n=0}^{\infty} \frac{(-1)^n 4^{n+2}}{(2n+1)!}
\]

To solve this, we can start by simplifying the general term and recognizing its structure. First, let’s factor out the \(4^2\) term to simplify:

\[
4^2 = 16, \quad \text{so we rewrite the series as} \quad 16 \sum_{n=0}^{\infty} \frac{(-1)^n 4^n}{(2n+1)!}
\]

This series looks similar to the Maclaurin series expansion for \(\sin(x)\), which is:

\[
\sin(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}
\]

In our case, we have powers of \(4\) instead of \(x\). Comparing the two, we see that the sum can be written as:

\[
\sin(2) = \sum_{n=0}^{\infty} \frac{(-1)^n 2^{2n+1}}{(2n+1)!}
\]

To match the structure, replace \(2\) with \(4\):

\[
16 \cdot \sin(4)
\]

Thus, the sum of the given series is:

\[
16 \cdot \sin(4)
\]

You can compute the final value of \(16 \sin(4)\) using a calculator to get a numerical result.

Let me know if you'd like further details or a step-by-step breakdown!

Here are 5 related questions to explore:
1. How does the Maclaurin series expansion for \(\cos(x)\) differ from \(\sin(x)\)?
2. Can you generalize this series for higher powers of 4?
3. What other common functions have similar Taylor or Maclaurin expansions?
4. How does the factorial term in the denominator affect convergence?
5. What is the radius of convergence for a typical sine series?

**Tip:** Recognizing patterns in series, such as the sine or cosine expansions, is a useful skill when working with infinite sums!

Find the sum of the series: ∑ (−1)^n 4^(n+2) / (2n+1)! from n=0 to infinity.

Learn how to solve the infinite series ∑ (−1)^n 4^(n+2) / (2n+1)! using the Maclaurin series expansion for sin(x). The detailed step-by-step solution involves recognizing the sine function's expansion and calculating 16 * sin(4). Suitable for undergraduate mathematics students.

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