Find the sum of the first n terms of the series:
`0.2 + 0.22 + 0.222+ ....n - terms`

Answer

To find the sum of the first n terms of the series:

\[ S_n = 0.2 + 0.22 + 0.222 + \ldots \text{ (n terms)} \]

we can follow these steps:

### Step 1: Rewrite the series
We can express each term in the series in a more manageable form. Notice that:

- \(0.2 = \frac{2}{10}\)
- \(0.22 = \frac{22}{100} = \frac{2 \times 11}{100}\)
- \(0.222 = \frac{222}{1000} = \frac{2 \times 111}{1000}\)

Thus, we can factor out 2 from each term:

\[ S_n = 2 \left( 0.1 + 0.11 + 0.111 + \ldots \right) \]

### Step 2: Express the inner series
The inner series can be expressed as:

\[ 0.1 + 0.11 + 0.111 + \ldots = 0.1 + \frac{11}{100} + \frac{111}{1000} + \ldots \]

### Step 3: Rewrite the inner series in terms of a geometric series
Notice that:

\[ 0.1 = \frac{1}{10}, \quad 0.11 = \frac{1}{10} + \frac{1}{100}, \quad 0.111 = \frac{1}{10} + \frac{1}{100} + \frac{1}{1000} \]

This can be rewritten as:

\[ 0.1 + 0.11 + 0.111 + \ldots = \sum_{k=1}^{n} \frac{1}{10^k} \]

### Step 4: Sum the geometric series
The sum of a geometric series can be calculated using the formula:

\[ S = a \frac{1 - r^n}{1 - r} \]

where \(a\) is the first term, \(r\) is the common ratio, and \(n\) is the number of terms. Here, \(a = \frac{1}{10}\), \(r = \frac{1}{10}\), and there are \(n\) terms.

Thus, we have:

\[ S_n = \frac{1/10 \left(1 - (1/10)^n\right)}{1 - 1/10} = \frac{1/10 \left(1 - (1/10)^n\right)}{9/10} = \frac{1 - (1/10)^n}{9} \]

### Step 5: Combine results
Now substituting back into our expression for \(S_n\):

\[ S_n = 2 \left( \frac{1 - (1/10)^n}{9} \right) = \frac{2(1 - (1/10)^n)}{9} \]

### Final Result
Thus, the sum of the first n terms of the series is:

\[ S_n = \frac{2}{9} \left( n - \frac{1 - (1/10)^n}{9} \right) \]