A : `(1)/(2)-(1)/(2).(1)/(2^(2))+(1)/(3).(1)/(2^(3))-(1)/(4).(1)/(2^(4))+....=log_(e)((3)/(2))`
R : `log_(e)(1+x)=x-(x^(2))/(2)+(x^(3))/(3)-(x^(4))/(4)+...`

Options

A. A is true, R is true and R is correct explanation of A
B. A is true, R is true and R is not correct explanation of A
C. A is true, R is false
D. A is false, R is true

Answer

To solve the problem, we need to analyze the series given in assertion A and the series given in reason R. We will show that the series in A is indeed equal to \( \log_e \left( \frac{3}{2} \right) \) by using the series expansion provided in R.

### Step-by-Step Solution:

1. **Understanding Assertion A:**
The series given in A is:
\[ A = \frac{1}{2} - \frac{1}{2} \cdot \frac{1}{2^2} + \frac{1}{3} \cdot \frac{1}{2^3} - \frac{1}{4} \cdot \frac{1}{2^4} + \ldots \]
This can be rewritten as:
\[ A = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n \cdot 2^n} \]

2. **Understanding Reason R:**
The series given in R is the Taylor series expansion for \( \log_e(1+x) \):
\[ \log_e(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots \]
We will substitute \( x = \frac{1}{2} \) into this series.

3. **Substituting \( x = \frac{1}{2} \) in R:**
\[ \log_e(1 + \frac{1}{2}) = \log_e(\frac{3}{2}) \]
Now substituting \( x = \frac{1}{2} \) into the series:
\[ \log_e(1 + \frac{1}{2}) = \frac{1}{2} - \frac{(\frac{1}{2})^2}{2} + \frac{(\frac{1}{2})^3}{3} - \frac{(\frac{1}{2})^4}{4} + \ldots \]
This simplifies to:
\[ \log_e(\frac{3}{2}) = \frac{1}{2} - \frac{1}{8} + \frac{1}{24} - \frac{1}{64} + \ldots \]

4. **Comparing the Two Series:**
Notice that the series we derived from R matches the series in A:
\[ A = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n \cdot 2^n} = \log_e(\frac{3}{2}) \]
Thus, we have shown that assertion A is indeed equal to \( \log_e \left( \frac{3}{2} \right) \).

5. **Conclusion:**
Since both assertion A and reason R are true, we conclude that:
\[ A = \log_e \left( \frac{3}{2} \right) \quad \text{and} \quad R \text{ is also true.} \]