To find the Maclaurin series for the function $f(x) = \ln(2 + x)$, we need to express this function as a power series around $x = 0$.

Step-by-step Solution:

1. Recall the Maclaurin series formula for $\ln(1 + u)$:

   The Maclaurin series for $\ln(1 + u)$ is: $\ln(1 + u) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} u^n, \quad \text{for } |u| < 1.$

   We want to rewrite $\ln(2 + x)$ in a form that makes it easier to use this formula.

2. Rewrite $\ln(2 + x)$:

   Set $u = \frac{x}{2}$, then we have: $\ln(2 + x) = \ln \left(2 \left(1 + \frac{x}{2}\right) \right).$ Using the logarithmic property $\ln(ab) = \ln a + \ln b$, this becomes: $\ln(2 + x) = \ln 2 + \ln \left(1 + \frac{x}{2}\right).$

   Now, we can apply the Maclaurin series expansion for $\ln(1 + u)$ with $u = \frac{x}{2}$:

   $\ln(1 + \frac{x}{2}) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \left(\frac{x}{2}\right)^n.$

3. Simplify the series:

   Substitute this series back into the expression for $\ln(2 + x)$:

   $\ln(2 + x) = \ln 2 + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \left(\frac{x}{2}\right)^n.$

   This can be written as:

   $\ln(2 + x) = \ln 2 + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \frac{x^n}{2^n}.$

4. Rewrite the series to match the provided options:

   We have:

   $\ln(2 + x) = \ln 2 - \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \frac{x^n}{2^n}.$

   This is equivalent to:

   $\ln(2 + x) = \ln 2 - \sum_{n=1}^{\infty} \frac{(-1)^n x^n}{n \cdot 2^n}.$

Conclusion:

This matches with Option D:

$\textbf{D. } \ln 2 - \sum_{n=1}^{\infty} \frac{(-1)^n x^n}{n \cdot 2^n}.$

Final Answer:

The correct answer is (d).

Would you like further details or explanations? Here are 5 related questions you might find interesting:

1. How do you derive the Maclaurin series for other logarithmic functions like $\ln(1 + x)$?
2. What is the radius of convergence for the series representation of $\ln(2 + x)$?
3. How does the series expansion change for higher derivatives of $\ln(2 + x)$?
4. Can you use the Maclaurin series to approximate $\ln(2.1)$?
5. What are the applications of logarithmic series in solving integrals?

Tip: For series convergence, always check the interval of convergence, especially when substituting variables or using logarithmic expansions.