IF `1+10+10^2+…….`upto n terms `=(10^n-1)/9` then the sum of the series 4+44+444+….upto n terms is

Options

A. `4/9(10^n-1)-(4n)/9`
B. `4/81(10^n-1)-(4n)/9`
C. `40/81(10^n-1)-(4n)/9`
D. `40/9(10^n-1)-(4n)/9`

Answer

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

### Step 1: Rewrite the series
The series can be rewritten by factoring out the common term \(4\):
\[ 4 + 44 + 444 + \ldots = 4(1 + 11 + 111 + \ldots) \]

### Step 2: Express the inner series
Now, we need to express the series \(1 + 11 + 111 + \ldots\) in a more manageable form. Notice that:
\[ 1 = 1 \times 10^0, \quad 11 = 1 \times 10^1 + 1 \times 10^0, \quad 111 = 1 \times 10^2 + 1 \times 10^1 + 1 \times 10^0 \]
Thus, we can express the \(k\)-th term as:
\[ 111\ldots1 \text{ (k times)} = \frac{10^k - 1}{9} \]
So, the series can be rewritten as:
\[ 1 + 11 + 111 + \ldots + \text{(k times)} = \sum_{k=1}^{n} \frac{10^k - 1}{9} \]

### Step 3: Simplify the sum
We can break this sum into two parts:
\[ \sum_{k=1}^{n} \frac{10^k}{9} - \sum_{k=1}^{n} \frac{1}{9} \]
The first part is a geometric series:
\[ \sum_{k=1}^{n} 10^k = 10 + 10^2 + \ldots + 10^n = 10 \frac{10^n - 1}{10 - 1} = \frac{10^{n+1} - 10}{9} \]
The second part is simply:
\[ \sum_{k=1}^{n} 1 = n \]
Thus, we have:
\[ \sum_{k=1}^{n} \frac{10^k - 1}{9} = \frac{1}{9} \left( \frac{10^{n+1} - 10}{9} - n \right) \]

### Step 4: Combine the results
Now substituting back into our original factor of \(4\):
\[ 4 \left( \frac{1}{9} \left( \frac{10^{n+1} - 10}{9} - n \right) \right) = \frac{4}{9} \left( \frac{10^{n+1} - 10}{9} - n \right) \]

### Step 5: Final simplification
Distributing \(4/9\):
\[ = \frac{4(10^{n+1} - 10)}{81} - \frac{4n}{9} \]

### Conclusion
Thus, the sum of the series \(4 + 44 + 444 + \ldots\) up to \(n\) terms is:
\[ \frac{4(10^{n+1} - 10)}{81} - \frac{4n}{9} \]