The sum of the series `1+(1)/(4)_(1)/(16)+(1)/(64)+....oo` is

Options

A. `(8)/(7)`
B. `(6)/(5)`
C. `(5)/(4)`
D. `(4)/(3)`

Explanation

To find the sum of the infinite geometric series \(1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \cdots\), follow these steps:

### Step-by-Step Solution:

1. **Identify the first term (a) and the common ratio (r):**
- The first term \(a\) of the series is \(1\).
- The common ratio \(r\) is the ratio of any term to its preceding term. Here, \(r = \frac{1}{4}\).

2. **Use the formula for the sum of an infinite geometric series:**
- The sum \(S\) of an infinite geometric series with first term \(a\) and common ratio \(r\) (where \(|r| < 1\)) is given by:
\[ S = \frac{a}{1 - r} \]

3. **Substitute the values of \(a\) and \(r\) into the formula:**
- Here, \(a = 1\) and \(r = \frac{1}{4}\).
\[ S = \frac{1}{1 - \frac{1}{4}} \]

4. **Simplify the expression:**
- Calculate \(1 - \frac{1}{4}\):
\[ 1 - \frac{1}{4} = \frac{4}{4} - \frac{1}{4} = \frac{3}{4} \]
- Substitute back into the formula:
\[ S = \frac{1}{\frac{3}{4}} = 1 \times \frac{4}{3} = \frac{4}{3} \]

### Final Answer:
The sum of the series \(1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \cdots\) is \(\frac{4}{3}\).