Find the sum of first n terms of:
`4+44 + 444+`…….

Answer

To find the sum of the first n terms of the series \(4 + 44 + 444 + \ldots\), we can follow these steps:

### Step 1: Identify the pattern in the series
The terms can be expressed as:
- 1st term: \(4\)
- 2nd term: \(44 = 4 \times 11\)
- 3rd term: \(444 = 4 \times 111\)

We can see that each term can be represented as \(4\) multiplied by a number that consists of \(n\) digits of \(1\).

### Step 2: Factor out the common term
We can factor out \(4\) from the series:
\[ S_n = 4 + 44 + 444 + \ldots + \text{(n terms)} = 4(1 + 11 + 111 + \ldots) \]

### Step 3: Rewrite the series inside the parentheses
The series inside the parentheses can be rewritten as:
\[ 1 + 11 + 111 + \ldots = 1 + (10 + 1) + (100 + 10 + 1) + \ldots \]
This can be expressed as:
\[ 1 + 11 + 111 + \ldots = \sum_{k=1}^{n} \frac{10^k - 1}{9} \]
This is because \(11 = \frac{10^2 - 1}{9}\), \(111 = \frac{10^3 - 1}{9}\), and so on.

### Step 4: Simplify the series
We can express the sum as:
\[ S_n = 4 \left( \sum_{k=1}^{n} \frac{10^k - 1}{9} \right) \]
This can be simplified to:
\[ S_n = \frac{4}{9} \left( \sum_{k=1}^{n} (10^k - 1) \right) \]

### Step 5: Calculate the sum of the geometric series
The sum of the geometric series \(10 + 10^2 + 10^3 + \ldots + 10^n\) can be calculated using the formula for the sum of a geometric series:
\[ \text{Sum} = a \frac{r^n - 1}{r - 1} \]
where \(a = 10\), \(r = 10\), and \(n\) is the number of terms. Thus:
\[ \sum_{k=1}^{n} 10^k = 10 \frac{10^n - 1}{10 - 1} = \frac{10}{9} (10^n - 1) \]

### Step 6: Substitute back into the equation
Substituting back, we have:
\[ S_n = \frac{4}{9} \left( \frac{10}{9} (10^n - 1) - n \right) \]

### Step 7: Final expression
Thus, the sum of the first \(n\) terms is:
\[ S_n = \frac{40}{81} (10^n - 1) - \frac{4n}{9} \]

### Final Answer
The sum of the first \(n\) terms of the series \(4 + 44 + 444 + \ldots\) is:
\[ S_n = \frac{40}{81} (10^n - 1) - \frac{4n}{9} \]