Find the sum of the series 0.6 +0.66 +0.666+ ... to the n terms

Answer

To find the sum of the series \( S_n = 0.6 + 0.66 + 0.666 + \ldots \) up to \( n \) terms, we can follow these steps:

### Step 1: Rewrite the Series
We can express each term in the series in a more manageable form:
\[ S_n = 0.6 + 0.66 + 0.666 + \ldots \]
This can be rewritten as:
\[ S_n = 0.6 + 0.66 + 0.666 + \ldots = 0.6 + 0.66 + 0.666 + \ldots + 0.6\underbrace{66\ldots6}_{n \text{ times}} \]

### Step 2: Factor Out Common Terms
Notice that each term can be expressed as:
\[ 0.6 = \frac{6}{10}, \quad 0.66 = \frac{66}{100} = \frac{6 \times 11}{100}, \quad 0.666 = \frac{666}{1000} = \frac{6 \times 111}{1000} \]
Thus, we can factor out \( 6 \) from the series:
\[ S_n = 6 \left( 0.1 + 0.11 + 0.111 + \ldots \right) \]

### Step 3: Express in Terms of a New Series
Now, we can express the series inside the parentheses:
\[ 0.1 + 0.11 + 0.111 + \ldots = 0.1(1 + 1 + 1 + \ldots) + 0.01(1 + 1 + \ldots) + 0.001(1 + 1 + \ldots) \]
This can be rewritten as:
\[ = 0.1 \sum_{k=0}^{n-1} 1 + 0.01 \sum_{k=0}^{n-2} 1 + 0.001 \sum_{k=0}^{n-3} 1 + \ldots \]

### Step 4: Use the Formula for the Sum of a Geometric Series
The series \( 0.1 + 0.01 + 0.001 + \ldots \) is a geometric series with the first term \( a = 0.1 \) and common ratio \( r = 0.1 \). The sum of the first \( n \) terms of a geometric series is given by:
\[ S = a \frac{1 - r^n}{1 - r} \]
Thus, we can write:
\[ S_n = 6 \left( \frac{0.1(1 - (0.1)^n)}{1 - 0.1} \right) = 6 \left( \frac{0.1(1 - 0.1^n)}{0.9} \right) \]

### Step 5: Simplify the Expression
Now, simplifying the expression gives:
\[ S_n = 6 \cdot \frac{0.1(1 - 0.1^n)}{0.9} = \frac{6 \cdot 0.1(1 - 0.1^n)}{0.9} = \frac{0.6(1 - 0.1^n)}{0.9} = \frac{2}{3}(1 - 0.1^n) \]

### Final Result
Thus, the sum of the series up to \( n \) terms is:
\[ S_n = \frac{2}{3} n - \frac{2}{9}(1 - 0.1^n) \]