The sum (upto two decimal places) of the infinite series `(7)/(17)+(77)/(17^(2))+(777)/(17^(3))+…………` is

Options

A. 1.06
B. 2.06
C. 3.06
D. 4.06

Answer

To find the sum of the infinite series \( S = \frac{7}{17} + \frac{77}{17^2} + \frac{777}{17^3} + \ldots \), we can follow these steps:

### Step 1: Identify the series structure
The series can be expressed as:
\[ S = \frac{7}{17} + \frac{77}{17^2} + \frac{777}{17^3} + \ldots \]
We can see that the numerators are formed by repeating the digit '7'.

### Step 2: Rewrite the numerators
Notice that:
- \( 7 = 7 \)
- \( 77 = 7 \times 11 \)
- \( 777 = 7 \times 111 \)

In general, the \( n \)-th term can be expressed as:
\[ \text{Numerator} = 7 \times (10^n - 1)/9 \]
This gives us:
\[ S = \frac{7}{17} + \frac{7 \times 11}{17^2} + \frac{7 \times 111}{17^3} + \ldots \]

### Step 3: Factor out the common term
We can factor out \( 7 \):
\[ S = 7 \left( \frac{1}{17} + \frac{11}{17^2} + \frac{111}{17^3} + \ldots \right) \]

### Step 4: Set up a new series
Let:
\[ T = \frac{1}{17} + \frac{11}{17^2} + \frac{111}{17^3} + \ldots \]
Now, we can express this series \( T \) in a more manageable form.

### Step 5: Multiply \( T \) by \( \frac{1}{17} \)
Multiply \( T \) by \( \frac{1}{17} \):
\[ \frac{T}{17} = \frac{1}{17^2} + \frac{11}{17^3} + \frac{111}{17^4} + \ldots \]

### Step 6: Subtract the two series
Now, subtract \( \frac{T}{17} \) from \( T \):
\[ T - \frac{T}{17} = \left( \frac{1}{17} + \frac{11}{17^2} + \frac{111}{17^3} + \ldots \right) - \left( \frac{1}{17^2} + \frac{11}{17^3} + \frac{111}{17^4} + \ldots \right) \]
This simplifies to:
\[ T - \frac{T}{17} = \frac{1}{17} + \left( \frac{11 - 1}{17^2} \right) + \left( \frac{111 - 11}{17^3} \right) + \ldots \]
This gives:
\[ T - \frac{T}{17} = \frac{1}{17} + \frac{10}{17^2} + \frac{100}{17^3} + \ldots \]

### Step 7: Recognize the new series as a geometric series
The new series \( \frac{1}{17} + \frac{10}{17^2} + \frac{100}{17^3} + \ldots \) is a geometric series with first term \( \frac{1}{17} \) and common ratio \( \frac{10}{17} \):
\[ \text{Sum} = \frac{\frac{1}{17}}{1 - \frac{10}{17}} = \frac{\frac{1}{17}}{\frac{7}{17}} = \frac{1}{7} \]

### Step 8: Solve for \( T \)
Now we can substitute back to find \( T \):
\[ T - \frac{T}{17} = \frac{1}{7} \]
This implies:
\[ \frac{16T}{17} = \frac{1}{7} \implies T = \frac{17}{16 \times 7} = \frac{17}{112} \]

### Step 9: Substitute back to find \( S \)
Now substituting \( T \) back into \( S \):
\[ S = 7T = 7 \times \frac{17}{112} = \frac{119}{112} \]

### Step 10: Convert to decimal
Now we convert \( \frac{119}{112} \) to decimal:
\[ \frac{119}{112} \approx 1.064 \]
Rounding to two decimal places gives:
\[ S \approx 1.06 \]

### Final Answer
The sum of the infinite series is approximately \( \boxed{1.06} \).