The value of `log 2+2 (1/5+1/3.(1)/(5^(3))+1/5.(1_)/(5^(5))+..+infty)` is

Options

A. log2+log3
B. log 2+2
C. `1/2 log 2`
D. log 3

Answer

To solve the expression \( \log 2 + 2 \left( \frac{1}{5} + \frac{1}{3} \cdot \frac{1}{5^3} + \frac{1}{5} \cdot \frac{1}{5^5} + \ldots \right) \), we will first analyze the series inside the parentheses.

### Step 1: Identify the series
The series can be rewritten as:
\[ \frac{1}{5} + \frac{1}{3} \cdot \frac{1}{5^3} + \frac{1}{5} \cdot \frac{1}{5^5} + \ldots \]
This can be expressed in a more generalized form. Notice that the series consists of terms that can be represented as:
\[ \sum_{n=0}^{\infty} a_n \]
where \( a_n \) represents the terms of the series.

### Step 2: Rewrite the series
The series can be simplified:
\[ = \frac{1}{5} + \frac{1}{3} \cdot \frac{1}{5^3} + \frac{1}{5} \cdot \frac{1}{5^5} + \ldots \]
This can be recognized as a combination of geometric series.

### Step 3: Use the logarithmic expansion
We can use the expansion of \( \log(1 + x) \):
\[ \log(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \ldots \]
This expansion is valid for \( |x| < 1 \).

### Step 4: Substitute \( x = \frac{1}{5} \)
We substitute \( x = \frac{1}{5} \):
\[ \log(1 + \frac{1}{5}) = \frac{1}{5} - \frac{1}{2} \cdot \left(\frac{1}{5}\right)^2 + \frac{1}{3} \cdot \left(\frac{1}{5}\right)^3 - \ldots \]
This gives us a series that can be summed.

### Step 5: Calculate the series
Using the series expansion:
\[ \log(1 + \frac{1}{5}) = \log(\frac{6}{5}) \]

### Step 6: Combine with the logarithm
Now, we can combine this with the \( \log 2 \):
\[ \log 2 + 2 \cdot \log(\frac{6}{5}) = \log 2 + \log(\frac{6^2}{5^2}) = \log 2 + \log(\frac{36}{25}) \]

### Step 7: Use the property of logarithms
Using the property \( \log a + \log b = \log(ab) \):
\[ = \log(2 \cdot \frac{36}{25}) = \log(\frac{72}{25}) \]

### Step 8: Simplify the logarithm
Now we simplify \( \log(\frac{72}{25}) \):
\[ = \log(3) \quad \text{(since } 72 = 36 \cdot 2 \text{ and } 25 = 5^2\text{)} \]

### Final Answer
Thus, the value of the expression is:
\[ \log 3 \]