The sum of the first n terms of the series `(1)/(2)+(3)/(4)+(7)/(8)+(15)/(16)+....` is equal to

Options

A. `2^(n)-n-1`
B. `1-2^(-n)`
C. `2^(-n)+n-1`
D. `2^(n)-1`

Explanation

To find the sum of the first n terms of the series \( \frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{15}{16} + \ldots \), we can follow these steps:

### Step 1: Identify the pattern in the series
The series can be rewritten as:
\[ \frac{1}{2} = 1 - \frac{1}{2}, \quad \frac{3}{4} = 1 - \frac{1}{4}, \quad \frac{7}{8} = 1 - \frac{1}{8}, \quad \frac{15}{16} = 1 - \frac{1}{16} \]
Thus, we can express the series as:
\[ \left(1 - \frac{1}{2}\right) + \left(1 - \frac{1}{4}\right) + \left(1 - \frac{1}{8}\right) + \left(1 - \frac{1}{16}\right) + \ldots \]

### Step 2: Separate the series into two parts
We can separate the series into two parts:
\[ \sum_{k=1}^{n} 1 - \sum_{k=1}^{n} \frac{1}{2^k} \]
The first part sums to \( n \) (since there are \( n \) terms of 1).

### Step 3: Calculate the sum of the geometric series
The second part is a geometric series:
\[ \sum_{k=1}^{n} \frac{1}{2^k} \]
The sum of the first \( n \) terms of a geometric series can be calculated using the formula:
\[ S_n = a \frac{1 - r^n}{1 - r} \]
where \( a = \frac{1}{2} \) and \( r = \frac{1}{2} \):
\[ S_n = \frac{1/2 \cdot (1 - (1/2)^n)}{1 - 1/2} = \frac{1/2 \cdot (1 - \frac{1}{2^n})}{\frac{1}{2}} = 1 - \frac{1}{2^n} \]

### Step 4: Combine the results
Now we can combine both parts:
\[ \text{Sum of first } n \text{ terms} = n - \left(1 - \frac{1}{2^n}\right) = n - 1 + \frac{1}{2^n} \]

### Final Answer
Thus, the sum of the first \( n \) terms of the series is:
\[ \boxed{n - 1 + \frac{1}{2^n}} \]