https://file.aiquickdraw.com/mathbot/file/1729931097390eyqwpf97.jpg

To solve this, we need to find the power series centered at 0 for the function \( f(x) = \frac{1}{(1 + 13x)^2} \). 

### Step-by-Step Solution
1. **Recognize the Structure of the Series**:
   The function is in the form \( f(x) = \frac{1}{(1 + 13x)^2} \), which suggests using a derivative approach from the geometric series \( \frac{1}{1 - u} = \sum_{n=0}^{\infty} u^n \) (valid for \( |u| < 1 \)).

2. **Rewrite Using a Derivative**:
   The hint suggests finding the derivative of \( \frac{1}{1 + 13x} \). Let:
   \[
   g(x) = \frac{1}{1 + 13x}
   \]
   so that \( f(x) = g'(x) \).

3. **Differentiate \( g(x) \)**:
   We know \( g(x) = \frac{1}{1 + 13x} \), which expands as:
   \[
   g(x) = \sum_{n=0}^{\infty} (-13x)^n = \sum_{n=0}^{\infty} (-1)^n 13^n x^n.
   \]
   To get \( f(x) = g'(x) \), differentiate term-by-term:
   \[
   f(x) = \frac{d}{dx} \left( \sum_{n=0}^{\infty} (-1)^n 13^n x^n \right) = \sum_{n=1}^{\infty} (-1)^n 13^n n x^{n-1}.
   \]

4. **Match the Result with the Given Choices**:
   This derived expression, \( f(x) = \sum_{n=1}^{\infty} (-1)^n 13^n n x^{n-1} \), matches the third answer option in the list.

### Answer:
The correct answer is:
\[
\sum_{n=1}^{\infty} (-1)^n 13^n n x^{n-1}.
\]

Would you like a detailed breakdown of each step or have questions on any part?

---

1. How is a power series expansion derived for a function like \( \frac{1}{(1 + 13x)^2} \)?
2. Why is the derivative approach useful in this context?
3. How do we determine the radius of convergence for such a series?
4. Can this method be applied to other functions of the form \( \frac{1}{(1 + ax)^2} \)?
5. What other techniques can we use to find power series for rational functions?

**Tip:** Remember that differentiating or integrating known power series expansions term-by-term can simplify finding series for more complex functions.

Find a power series centered at 0 for the function: f(x) = \( \frac{1}{(1 + 13x)^2} \). (Hint: what is the derivative of \( \frac{1}{1 + 13x} \)?)

Find the power series expansion centered at 0 for the function f(x) = \( \frac{1}{(1 + 13x)^2} \). Learn to apply geometric series and term-by-term differentiation to derive the answer, using foundational calculus techniques. Ideal for undergraduate calculus 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