https://file.aiquickdraw.com/mathbot/file/1729020738817ufwf9e7d.jpg

The given problem asks for the first three terms of the power series:

\[
\sum_{n=0}^{\infty} \frac{(3)^n}{n!} x^{n+4}
\]

To find the first three terms, we substitute \(n = 0\), \(n = 1\), and \(n = 2\) into the general formula.

### Step-by-step calculation:

1. **For \(n = 0\):**
   \[
   \frac{(3)^0}{0!} x^{0+4} = \frac{1}{1} x^4 = x^4
   \]

2. **For \(n = 1\):**
   \[
   \frac{(3)^1}{1!} x^{1+4} = \frac{3}{1} x^5 = 3x^5
   \]

3. **For \(n = 2\):**
   \[
   \frac{(3)^2}{2!} x^{2+4} = \frac{9}{2} x^6 = \frac{9}{2}x^6
   \]

### Therefore, the first three terms of the power series are:
\[
x^4 + 3x^5 + \frac{9}{2}x^6
\]

Let me know if you want further clarification or more details!

### Here are 5 related questions to deepen your understanding:
1. How do we determine the general term of a power series?
2. What is the radius of convergence for this particular power series?
3. How would the series change if the coefficient \(3^n\) were replaced with \(n^3\)?
4. How do factorial terms like \(n!\) affect the growth of a power series?
5. How does the exponent of \(x\) impact the behavior of the power series for large values of \(x\)?

**Tip:** When working with series, always remember to check for convergence using appropriate tests, such as the ratio test, if you're dealing with an infinite number of terms.

Write out the first 3 terms of the power series \( \sum_{n=0}^{\infty} \frac{(3)^n}{n!} x^{n+4} \).

Learn how to compute the first three terms of the power series \( \sum_{n=0}^{\infty} \frac{(3)^n}{n!} x^{n+4} \). The problem involves power series, factorial notation, and exponential growth. Suitable for undergraduate students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation