Find the sum of the geometric series : `1,(1)/(2),(1)/(4),(1)/(8), . . . .. . . . . .` upto 12 terms.

Answer

To find the sum of the geometric series \(1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \ldots\) up to 12 terms, we can follow these steps:

### Step 1: Identify the first term and common ratio
The first term \(a\) of the series is:
\[ a = 1 \]
The common ratio \(r\) can be found by dividing the second term by the first term:
\[ r = \frac{\frac{1}{2}}{1} = \frac{1}{2} \]

### Step 2: Confirm that the common ratio is less than 1
Since \(r = \frac{1}{2}\), which is less than 1, we can use the formula for the sum of the first \(n\) terms of a geometric series.

### Step 3: Use the formula for the sum of the first \(n\) terms
The formula for the sum \(S_n\) of the first \(n\) terms of a geometric series is:
\[ S_n = a \frac{1 - r^n}{1 - r} \]
For our case, we want to find \(S_{12}\):
\[ S_{12} = 1 \cdot \frac{1 - \left(\frac{1}{2}\right)^{12}}{1 - \frac{1}{2}} \]

### Step 4: Simplify the expression
Now, substituting the values into the formula:
\[ S_{12} = \frac{1 - \left(\frac{1}{2}\right)^{12}}{\frac{1}{2}} = 2 \left(1 - \left(\frac{1}{2}\right)^{12}\right) \]

### Step 5: Calculate \(\left(\frac{1}{2}\right)^{12}\)
Calculating \(\left(\frac{1}{2}\right)^{12}\):
\[ \left(\frac{1}{2}\right)^{12} = \frac{1}{4096} \]

### Step 6: Substitute back into the sum formula
Now substituting this back into our expression for \(S_{12}\):
\[ S_{12} = 2 \left(1 - \frac{1}{4096}\right) = 2 \left(\frac{4096 - 1}{4096}\right) = 2 \cdot \frac{4095}{4096} = \frac{8190}{4096} \]

### Step 7: Simplify the final result
The final result can be simplified:
\[ S_{12} = \frac{8190}{4096} \]

Thus, the sum of the geometric series up to 12 terms is:
\[ \boxed{\frac{8190}{4096}} \]