Let's solve this step by step to represent $f(x) = \frac{5}{3 - x}$ as a power series with center $x = -2$, filling in the blank box.

Step 1: Rewrite $f(x)$ to resemble a geometric series

The general formula for a geometric series is: $\frac{1}{1 - r} = \sum_{n=0}^\infty r^n, \quad \text{for } |r| < 1.$

We first rewrite $f(x)$ in a similar form: $f(x) = \frac{5}{3 - x}.$

We aim to express this in terms of $x + 2$ (centered at $x = -2$). Substitute $x = u - 2$, which makes $u = x + 2$: $f(x) = \frac{5}{3 - (-2 + u)} = \frac{5}{5 - u}.$

Factor out the $5$ in the denominator: $f(x) = \frac{5}{5(1 - \frac{u}{5})} = \frac{1}{1 - \frac{u}{5}}.$

Rewriting $u$ as $x + 2$, we have: $f(x) = \frac{5}{5(1 - \frac{x + 2}{5})} = \frac{5}{3 - x} = 5 \sum_{n=0}^\infty \left( \frac{x + 2}{5} \right)^n,$ valid when $|x + 2| < 5$.

Step 2: Fill in the blank

The power series representation of $f(x)$ becomes: $f(x) = 5 \sum_{n=0}^\infty \left( \frac{x + 2}{5} \right)^n = \sum_{n=0}^\infty \frac{5}{5^n} (x + 2)^n = \sum_{n=0}^\infty \frac{1}{5^{n-1}} (x + 2)^n.$

Thus, the box should contain: $\left( \frac{x + 2}{5} \right)^n$

Let me know if you'd like more clarification!

Further questions:

1. Why does the geometric series require $|r| < 1$?
2. How do we derive the radius of convergence $|x + 2| < 5$?
3. What happens if $x$ lies outside this interval?
4. How can you extend the power series representation to functions with $x^2$ terms?
5. What are some applications of power series representations?

Tip:

Always rewrite the denominator to match the geometric series form $\frac{1}{1 - r}$, factoring constants as needed!