1. **Problem Statement:** Find the Maclaurin series expansion for the function $f(x) = \frac{1}{x-2}$. 2. **Recall the Maclaurin series formula:** The Maclaurin series of a function $f(x)$ is given by $$f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n,$$ where $f^{(n)}(0)$ is the $n$th derivative of $f$ evaluated at 0. 3. **Rewrite the function:** To find the Maclaurin series, express $f(x)$ in a form suitable for expansion around $x=0$: $$f(x) = \frac{1}{x-2} = \frac{1}{-2 + x} = -\frac{1}{2 - x} = -\frac{1}{2(1 - \frac{x}{2})} = -\frac{1}{2} \cdot \frac{1}{1 - \frac{x}{2}}.$$ 4. **Use the geometric series expansion:** Recall that for $|r| < 1$, $$\frac{1}{1-r} = \sum_{n=0}^\infty r^n.$$ Here, $r = \frac{x}{2}$, so $$\frac{1}{1 - \frac{x}{2}} = \sum_{n=0}^\infty \left(\frac{x}{2}\right)^n = \sum_{n=0}^\infty \frac{x^n}{2^n}.$$ 5. **Combine the results:** Substitute back into $f(x)$: $$f(x) = -\frac{1}{2} \sum_{n=0}^\infty \frac{x^n}{2^n} = -\sum_{n=0}^\infty \frac{x^n}{2^{n+1}}.$$ 6. **Interpretation:** The Maclaurin series for $f(x) = \frac{1}{x-2}$ is $$- \sum_{n=0}^\infty \frac{x^n}{2^{n+1}}.$$ 7. **Answer choice:** This matches option C. **Final answer:** $$f(x) = - \sum_{n=0}^\infty \frac{x^n}{2^{n+1}}.$$