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.