To find a power series for the function $f(x) = \frac{1}{1 + 6x}$, we can start by recognizing that this function is similar to the standard geometric series formula:

$\frac{1}{1 - r} = 1 + r + r^2 + r^3 + \ldots$

This series converges when $|r| < 1$.

Rewriting $f(x)$

We can rewrite $f(x) = \frac{1}{1 + 6x}$ in a form that fits the geometric series formula by factoring out the negative sign in the denominator:

$f(x) = \frac{1}{1 - (-6x)}$

Here, the common ratio $r$ is $-6x$. Therefore, the power series expansion for $f(x)$ is:

$f(x) = 1 + (-6x) + (-6x)^2 + (-6x)^3 + \ldots$

Simplifying each term, we get:

$f(x) = 1 - 6x + 36x^2 - 216x^3 + \ldots + (-6)^n x^n + \ldots$

Matching with the given choices:

This matches with Choice D:

$\boxed{1 - 6x + 36x^2 + \ldots + (-6)^n x^n + \ldots}$

This is the correct power series representation for $f(x) = \frac{1}{1 + 6x}$.

Would you like more details or have any questions? Here are some related questions to explore:

1. What is the interval of convergence for the power series of $f(x) = \frac{1}{1 + 6x}$?
2. How do you find the power series expansion for more complex functions?
3. What happens to the series when $|6x| \geq 1$?
4. How would the power series change if the function were $f(x) = \frac{1}{1 - 6x}$?
5. Can you derive the power series for $f(x) = \frac{1}{1 + ax}$ for any constant $a$?

Tip: Power series are particularly useful for approximating functions near a specific point, often near zero, where they converge.