`-((4)/(5)+(4^(2))/(5^(2).2)+(4^(3))/(5^(3).3)+(4^(4))/(5^(4).4)+……oo)=`

Options

A. a) `"log"(9)/(5)`
B. b) `"log"(4)/(5)`
C. c) `log((5)/(4))`
D. d) `"log"(1)/(5)`

Answer

To solve the given series:

\[ - \left( \frac{4}{5} + \frac{4^2}{5^2 \cdot 2} + \frac{4^3}{5^3 \cdot 3} + \frac{4^4}{5^4 \cdot 4} + \ldots \right) \]

we can recognize that this series resembles the Taylor series expansion for the logarithm function. Let's break it down step-by-step.

### Step 1: Identify the series
The series can be rewritten as:

\[ - \sum_{n=1}^{\infty} \frac{4^n}{5^n \cdot n} \]

### Step 2: Factor out common terms
We can factor out \(\frac{4}{5}\) from the series:

\[ - \sum_{n=1}^{\infty} \frac{(4/5)^n}{n} \]

### Step 3: Recognize the series as a logarithm
The series \(\sum_{n=1}^{\infty} \frac{x^n}{n}\) is known to converge to \(-\log(1-x)\) for \(|x| < 1\). Here, we have \(x = \frac{4}{5}\):

\[ \sum_{n=1}^{\infty} \frac{(4/5)^n}{n} = -\log(1 - \frac{4}{5}) = -\log(\frac{1}{5}) \]

### Step 4: Substitute back into the expression
Now substituting this back into our expression, we have:

\[ - \left(-\log(\frac{1}{5})\right) = \log(\frac{1}{5}) \]

### Final Answer
Thus, the value of the original series is:

\[ \log(\frac{1}{5}) \]