To find the power series centered at 0 for the function $f(x) = \frac{1}{(1+13x)^2}$, we can proceed by following these steps:

1. Identify the Known Series Expansion: Start with the geometric series: $\frac{1}{1 - u} = \sum_{n=0}^{\infty} u^n \quad \text{for } |u| < 1.$

2. Rewrite the Function: Rewrite $f(x)$ in a form that resembles the geometric series. Let $u = -13x$, so that: $\frac{1}{1 + 13x} = \frac{1}{1 - (-13x)} = \sum_{n=0}^{\infty} (-13x)^n = \sum_{n=0}^{\infty} (-13)^n x^n.$

3. Differentiate to Get $\frac{1}{(1+13x)^2}$: Since our target function is $\frac{1}{(1+13x)^2}$, we differentiate the geometric series term by term with respect to $x$: $\frac{d}{dx} \left( \frac{1}{1 + 13x} \right) = \frac{d}{dx} \left( \sum_{n=0}^{\infty} (-13)^n x^n \right).$

   Differentiating the left side gives: $\frac{d}{dx} \left( \frac{1}{1 + 13x} \right) = -\frac{13}{(1 + 13x)^2}.$

   To match $f(x) = \frac{1}{(1 + 13x)^2}$, divide by $-13$: $\frac{1}{(1 + 13x)^2} = \sum_{n=1}^{\infty} n(-13)^n x^{n-1}.$

So the power series for $f(x) = \frac{1}{(1+13x)^2}$ is: $f(x) = \sum_{n=1}^{\infty} n(-13)^n x^{n-1}.$

Do you want more details on these steps or have questions?

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5 Related Questions:

1. How would the series change if the function were $\frac{1}{(1+ax)^2}$ with a different constant $a$?
2. How can we find the radius of convergence for this series?
3. What happens if we want a power series for $\frac{1}{(1+13x)^3}$ instead?
4. How does differentiation affect the convergence of a series?
5. Could we obtain this series by expanding $\frac{1}{1+13x}$ and then squaring it?

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One Tip:

To derive power series, rewriting the function in terms of simpler series forms, like the geometric series, often simplifies differentiation and integration.