`e^(2((1)/(3)+(1)/(3)*(1)/(3^(3))+(1)/(5)*(1)/(3^(5))+….))=`

Options

A. 2
B. 4
C. 1
D. 3

Answer

To solve the expression \( e^{2\left(\frac{1}{3} + \frac{1}{3} \cdot \frac{1}{3^3} + \frac{1}{5} \cdot \frac{1}{3^5} + \ldots\right)} \), we first need to analyze the series inside the exponent.

### Step 1: Identify the series
The series can be rewritten as:
\[ S = \frac{1}{3} + \frac{1}{3} \cdot \frac{1}{3^3} + \frac{1}{5} \cdot \frac{1}{3^5} + \ldots \]
We can observe that the terms are of the form:
\[ S = \sum_{n=1}^{\infty} \frac{1}{2n+1} \cdot \frac{1}{3^{2n-1}} \]

### Step 2: Recognize the series as a logarithmic function
The series resembles the Taylor series expansion for the logarithmic function. Specifically, we can relate it to the series:
\[ \ln(1+x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \ldots \]
We can rewrite our series in a form that allows us to use this logarithmic identity.

### Step 3: Transform the series
Notice that:
\[ S = \sum_{n=1}^{\infty} \frac{1}{2n+1} \left(\frac{1}{3}\right)^{2n-1} \]
This can be transformed into:
\[ S = \frac{1}{2} \ln\left(\frac{1}{1 - \frac{1}{3}}\right) = \frac{1}{2} \ln\left(\frac{3}{2}\right) \]

### Step 4: Substitute back into the exponent
Now substituting \( S \) back into the exponent:
\[ e^{2S} = e^{2 \cdot \frac{1}{2} \ln\left(\frac{3}{2}\right)} = e^{\ln\left(\frac{3}{2}\right)} \]

### Step 5: Simplify the expression
Using the property of exponents and logarithms, we have:
\[ e^{\ln\left(\frac{3}{2}\right)} = \frac{3}{2} \]

### Final Result
Thus, the final result is:
\[ \frac{3}{2} \]