The task is to expand the given function $e^{-2x - 1}$ into a Maclaurin series up to terms of $O(x^n)$.

Step-by-step explanation:

1. The given function: $f(x) = e^{-2x - 1}$ The Maclaurin series for $e^u$ is: $e^u = \sum_{k=0}^\infty \frac{u^k}{k!}, \quad u = -2x - 1$ For the Maclaurin series, we substitute $u = -2x - 1$.

2. Expansion:
   Substitute $u = -2x - 1$ into the series: $e^{-2x - 1} = \sum_{k=0}^\infty \frac{(-2x - 1)^k}{k!}$

   Now expand the first few terms explicitly, up to $O(x^n)$, based on the problem's request.

3. Simplify for lower powers of $x$: Expand $(-2x - 1)^k$ using the Binomial Theorem: $(-2x - 1)^k = \sum_{j=0}^k \binom{k}{j}(-2x)^j(-1)^{k-j}$ Substitute into the series for $e^{-2x - 1}$.

4. Extract terms up to the desired order $O(x^n)$: Each term of the series is calculated as: $\frac{(-2x - 1)^k}{k!} = \frac{1}{k!} \sum_{j=0}^k \binom{k}{j} (-2x)^j (-1)^{k-j}$ Combine terms of equal power of $x$ and stop at the requested order $O(x^n)$.

Would you like a detailed expansion of this step with a specific $n$ (e.g., $n = 3$ or $n = 4$)? Let me know!