The sum of the sequence upto 5, 55, 555, …. upto n infinite terms is

Options

A. `(5)/(9) [(10(10^(n)-1)+n)/(9)]`
B. `(5)/(9) [(10(10^(n)-1))/(9)-n]`
C. `(5)/(9) [(10(10^(npm1)-1))/(9)-n]`
D. `(5)/(9) [(10(10^(n-1)-1))/(9)-n]`

Answer

To find the sum of the sequence \(5, 55, 555, \ldots\) up to \(n\) infinite terms, we can follow these steps:

### Step 1: Identify the Sequence
The sequence can be expressed as:
- \(5 = 5\)
- \(55 = 5 \times 11\)
- \(555 = 5 \times 111\)

Thus, the sequence can be rewritten as:
\[ S = 5 + 55 + 555 + \ldots \]

### Step 2: Factor Out the Common Term
We can factor out \(5\) from the sum:
\[ S = 5(1 + 11 + 111 + \ldots) \]

### Step 3: Rewrite the Inner Series
The inner series can be expressed in terms of powers of \(10\):
- \(1 = 1\)
- \(11 = 10 + 1\)
- \(111 = 100 + 10 + 1\)

So, we can express the series as:
\[ 1 + 11 + 111 + \ldots = 1 + (10 + 1) + (100 + 10 + 1) + \ldots \]

### Step 4: Express the Inner Series as a Geometric Series
The series can be rewritten as:
\[ 1 + 10 + 100 + \ldots \]
This is a geometric series where:
- First term \(a = 1\)
- Common ratio \(r = 10\)

### Step 5: Use the Formula for the Sum of an Infinite Geometric Series
The sum of an infinite geometric series is given by:
\[ S = \frac{a}{1 - r} \]
For our series:
\[ S = \frac{1}{1 - 10} = \frac{1}{-9} \]

### Step 6: Combine the Results
Now substituting back into our expression for \(S\):
\[ S = 5 \left( \frac{1}{-9} \right) = -\frac{5}{9} \]

### Step 7: Final Result
Thus, the sum of the sequence up to \(n\) infinite terms is:
\[ S = -\frac{5}{9} \]