The sum of 4 + 44 + 444 + ............... .. up to 100 terms.

Options

A. `4/9[(10(10^100-1))/9-100]`
B. `4/9[(10(10^100-1))/9-1]`
C. `4/81[(10(10^100-1))/9-1]`
D. `2/81[(10(10^100-1))/9-100]`

Answer

To solve the problem of finding the sum of the series \(4 + 44 + 444 + \ldots\) up to 100 terms, we can break it down into manageable steps.

### Step-by-Step Solution:

1. **Identify the Pattern**:
The terms of the series can be represented as:
- 1st term: \(4\)
- 2nd term: \(44 = 4 \times 11\)
- 3rd term: \(444 = 4 \times 111\)
- 4th term: \(4444 = 4 \times 1111\)

We can see that each term can be expressed as \(4\) multiplied by a number consisting of repeated \(1\)s.

2. **Express Each Term**:
The \(n\)-th term can be expressed as:
\[ T_n = 4 \times (10^n - 1)/9 \]
This is because \(111...1\) (with \(n\) ones) can be represented as \((10^n - 1)/9\).

3. **Sum of the Series**:
The sum \(S\) of the first 100 terms can be expressed as:
\[ S = 4 \left(1 + 11 + 111 + \ldots + \text{(100 terms)}\right) \]

4. **Using the Formula for the Sum of a Geometric Series**:
The series inside the parentheses can be rewritten as:
\[ S = 4 \left(\sum_{n=1}^{100} \frac{10^n - 1}{9}\right) \]
This can be simplified to:
\[ S = \frac{4}{9} \left(10 + 10^2 + 10^3 + \ldots + 10^{100} - 100\right) \]

5. **Calculate the Sum of the Geometric Series**:
The sum of the geometric series \(10 + 10^2 + 10^3 + \ldots + 10^{100}\) can be calculated using the formula for the sum of a geometric series:
\[ S_n = a \frac{r^n - 1}{r - 1} \]
where \(a = 10\), \(r = 10\), and \(n = 100\):
\[ S = 10 \frac{10^{100} - 1}{10 - 1} = \frac{10^{101} - 10}{9} \]

6. **Final Calculation**:
Substitute back into the sum:
\[ S = \frac{4}{9} \left(\frac{10^{101} - 10}{9} - 100\right) \]
Simplifying this gives us:
\[ S = \frac{4}{81} (10^{101} - 10 - 900) \]
\[ S = \frac{4}{81} (10^{101} - 910) \]

### Final Answer:
Thus, the sum of the series \(4 + 44 + 444 + \ldots\) up to 100 terms is:
\[ S = \frac{4}{81} (10^{101} - 910) \]