What is the sum of the series 0.5+0.55+0.555+. . .+n terms ?

Options

A. `(5)/(9)[n-(2)/(9)(1-(1)/(10^(n)))]`
B. `(1)/(9)[5-(2)/(9)(1-(1)/(10^(n)))]`
C. `(1)/(9)[n-(5)/(9)(1-(1)/(10^(n)))]`
D. `(5)/(9)[n-(1)/(9)(1-(1)/(10^(n)))]`

Answer

To find the sum of the series \( S = 0.5 + 0.55 + 0.555 + \ldots \) for \( n \) terms, we can follow these steps:

### Step 1: Rewrite the Series
The terms of the series can be expressed in a more manageable form. Each term can be represented as:
- \( 0.5 = \frac{5}{10} \)
- \( 0.55 = \frac{55}{100} = \frac{5 \times 11}{10^2} \)
- \( 0.555 = \frac{555}{1000} = \frac{5 \times 111}{10^3} \)

Thus, the \( k \)-th term can be written as:
\[ \text{Term}_k = \frac{5 \times (10^k - 1)/9}{10^k} \]
This simplifies to:
\[ \text{Term}_k = \frac{5}{9} \left(1 - \frac{1}{10^k}\right) \]

### Step 2: Sum the Series
The series can be rewritten as:
\[ S_n = \sum_{k=1}^{n} \frac{5}{9} \left(1 - \frac{1}{10^k}\right) \]
This can be separated into two parts:
\[ S_n = \frac{5}{9} \left( \sum_{k=1}^{n} 1 - \sum_{k=1}^{n} \frac{1}{10^k} \right) \]
The first sum is simply \( n \):
\[ \sum_{k=1}^{n} 1 = n \]
The second sum is a geometric series:
\[ \sum_{k=1}^{n} \frac{1}{10^k} = \frac{\frac{1}{10}(1 - (\frac{1}{10})^n)}{1 - \frac{1}{10}} = \frac{1}{9} \left(1 - \frac{1}{10^n}\right) \]

### Step 3: Substitute Back
Now substituting back into the equation for \( S_n \):
\[ S_n = \frac{5}{9} \left( n - \frac{1}{9} \left(1 - \frac{1}{10^n}\right) \right) \]
This simplifies to:
\[ S_n = \frac{5n}{9} - \frac{5}{81} \left(1 - \frac{1}{10^n}\right) \]

### Step 4: Final Expression
Thus, the final expression for the sum of the series is:
\[ S_n = \frac{5n}{9} - \frac{5}{81} + \frac{5}{81 \cdot 10^n} \]

### Summary
The sum of the series \( 0.5 + 0.55 + 0.555 + \ldots \) for \( n \) terms is:
\[ S_n = \frac{5n}{9} - \frac{5}{81} + \frac{5}{81 \cdot 10^n} \]