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 pattern in the series
The series can be rewritten in a more manageable form. Notice that:
- The first term is \( 7 \)
- The second term is \( 77 = 7 \times 11 \)
- The third term is \( 777 = 7 \times 111 \)

We can express each term as:
\[ \frac{7 \times (10^n - 1)/9}{17^n} \quad \text{for } n = 1, 2, 3, \ldots \]
where \( 10^n - 1 \) generates the sequence of numbers composed of \( n \) digits of \( 7 \).

### Step 2: Rewrite the series
Thus, we can express the series as:
\[ S = \sum_{n=1}^{\infty} \frac{7 \times (10^n - 1)/9}{17^n} \]
This can be simplified to:
\[ S = \frac{7}{9} \sum_{n=1}^{\infty} \left( \frac{10^n}{17^n} - \frac{1}{17^n} \right) \]

### Step 3: Split the series
We can split the series into two parts:
\[ S = \frac{7}{9} \left( \sum_{n=1}^{\infty} \left( \frac{10}{17} \right)^n - \sum_{n=1}^{\infty} \left( \frac{1}{17} \right)^n \right) \]

### Step 4: Calculate the sums of the geometric series
Both sums are geometric series. The sum of a geometric series \( \sum_{n=0}^{\infty} ar^n \) is given by \( \frac{a}{1 - r} \) where \( |r| < 1 \).

1. For the first sum:
\[ \sum_{n=1}^{\infty} \left( \frac{10}{17} \right)^n = \frac{\frac{10}{17}}{1 - \frac{10}{17}} = \frac{\frac{10}{17}}{\frac{7}{17}} = \frac{10}{7} \]

2. For the second sum:
\[ \sum_{n=1}^{\infty} \left( \frac{1}{17} \right)^n = \frac{\frac{1}{17}}{1 - \frac{1}{17}} = \frac{\frac{1}{17}}{\frac{16}{17}} = \frac{1}{16} \]

### Step 5: Substitute back into the equation for \( S \)
Now substituting these back into the equation for \( S \):
\[ S = \frac{7}{9} \left( \frac{10}{7} - \frac{1}{16} \right) \]

### Step 6: Simplify the expression
To simplify \( \frac{10}{7} - \frac{1}{16} \), we need a common denominator:
\[ \frac{10}{7} = \frac{160}{112}, \quad \frac{1}{16} = \frac{7}{112} \]
Thus,
\[ \frac{10}{7} - \frac{1}{16} = \frac{160}{112} - \frac{7}{112} = \frac{153}{112} \]

### Step 7: Final calculation of \( S \)
Now substituting this back:
\[ S = \frac{7}{9} \cdot \frac{153}{112} = \frac{1071}{1008} \]

### Step 8: Convert to decimal
Calculating \( \frac{1071}{1008} \) gives approximately \( 1.06 \) when rounded to two decimal places.

### Final Answer
Thus, the sum of the infinite series up to two decimal places is:
\[ \boxed{1.06} \]