**Find the Maclaurin series for the function**

**f(x)=ln(2+x).**

**Question 2**

**Select one:**




**a.**

**ln2−**

**∑**

**∞**

**n=1**

**x**

**n**

**n**

**2**

**n**







**b.**

**ln2−**

**∑**

**∞**

**n=1**

**x**

**n**

**n**







**c.**

**ln2−**

**∑**

**∞**

**n=1**

**(−1**

**)**

**n**

**x**

**n/**

**n**







**d.**

**ln2−**

**∑**

**∞**

**n=1**

**(−1**

**)**

**n**

**x**

**n/**

**n**

**2**

**n**

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.

Explore the Maclaurin series for ln(2 + x) with a detailed step-by-step solution. Learn how to derive and apply the Maclaurin series formula for logarithmic functions. Suitable for college-level mathematics.

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