`(2)/(1). (1)/(3) + (3)/(2) . (1)/(9) + (4)/(3).(1)/(27) + (5)/(4).(1)/(81) + ... =` ..........

Answer

To solve the series

\[ \frac{2}{1} \cdot \frac{1}{3} + \frac{3}{2} \cdot \frac{1}{9} + \frac{4}{3} \cdot \frac{1}{27} + \frac{5}{4} \cdot \frac{1}{81} + \ldots \]

we will first identify the general term of the series.

### Step 1: Identify the General Term
The series can be expressed in terms of a summation. Observing the pattern, we can see that:

- The numerator of each term increases by 1 starting from 2: \(2, 3, 4, 5, \ldots\) which can be represented as \(n + 1\) where \(n\) starts from 1.
- The denominator of the first part is \(1, 2, 3, 4, \ldots\) which is simply \(n\).
- The second part of each term is a power of \(\frac{1}{3}\): \(\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \frac{1}{81}, \ldots\) which can be represented as \(\left(\frac{1}{3}\right)^n\).

Thus, the general term \(T_n\) can be written as:

\[ T_n = \frac{n + 1}{n} \cdot \left(\frac{1}{3}\right)^n \]

### Step 2: Rewrite the Series
We can express the series as:

\[ S = \sum_{n=1}^{\infty} \frac{n + 1}{n} \cdot \left(\frac{1}{3}\right)^n \]

This can be separated into two parts:

\[ S = \sum_{n=1}^{\infty} \left(1 + \frac{1}{n}\right) \left(\frac{1}{3}\right)^n = \sum_{n=1}^{\infty} \left(\frac{1}{3}\right)^n + \sum_{n=1}^{\infty} \frac{1}{n} \left(\frac{1}{3}\right)^n \]

### Step 3: Calculate Each Part
1. **First Part**:

The first part is a geometric series:

\[ \sum_{n=1}^{\infty} \left(\frac{1}{3}\right)^n = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2} \]

2. **Second Part**:

The second part can be recognized as the series for \(-\log(1 - x)\) where \(x = \frac{1}{3}\):

\[ \sum_{n=1}^{\infty} \frac{1}{n} \left(\frac{1}{3}\right)^n = -\log\left(1 - \frac{1}{3}\right) = -\log\left(\frac{2}{3}\right) = \log\left(\frac{3}{2}\right) \]

### Step 4: Combine Results
Now we can combine the results from the two parts:

\[ S = \frac{1}{2} + \log\left(\frac{3}{2}\right) \]

### Step 5: Final Simplification
Using the property of logarithms, we can express this as:

\[ S = \frac{1}{2} + \log 3 - \log 2 = \frac{1}{2} + \log\left(\frac{3}{2}\right) \]

### Final Answer
Thus, the sum of the series is:

\[ \frac{1}{2} + \log 3 - \log 2 \]