The sum of the first n terms of the series `(1)/(2)+(3)/(4)+(7)/(8)+(15)/(16)+`. . . . Is equal to
(a)`2^(n)-n-1`
(b)`1-2^(-n)`
(c)`2^(-n)+n-1`
(d)`2^(n)-1`

Options

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

Answer

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 analyze the pattern of the series and check the given options.

### Step-by-Step Solution:

1. **Identify the Pattern in the Series**:
The numerators of the series are \( 1, 3, 7, 15, \ldots \). We can observe that:
- \( 1 = 2^1 - 1 \)
- \( 3 = 2^2 - 1 \)
- \( 7 = 2^3 - 1 \)
- \( 15 = 2^4 - 1 \)

Thus, the pattern for the numerator can be generalized as \( 2^n - 1 \).

2. **Identify the Denominators**:
The denominators of the series are \( 2, 4, 8, 16, \ldots \). This can be expressed as:
- \( 2^1, 2^2, 2^3, 2^4, \ldots \)

Therefore, the general term of the series can be written as:
\[ \frac{2^n - 1}{2^n} \]

3. **Sum of the First n Terms**:
The sum of the first n terms \( S_n \) can be expressed as:
\[ S_n = \sum_{k=1}^{n} \frac{2^k - 1}{2^k} \]
This can be split into two separate sums:
\[ S_n = \sum_{k=1}^{n} \left( 1 - \frac{1}{2^k} \right) = \sum_{k=1}^{n} 1 - \sum_{k=1}^{n} \frac{1}{2^k} \]
The first sum \( \sum_{k=1}^{n} 1 \) equals \( n \).

4. **Calculate the Second Sum**:
The second sum \( \sum_{k=1}^{n} \frac{1}{2^k} \) is a geometric series with first term \( \frac{1}{2} \) and common ratio \( \frac{1}{2} \):
\[ \sum_{k=1}^{n} \frac{1}{2^k} = \frac{\frac{1}{2}(1 - (\frac{1}{2})^n)}{1 - \frac{1}{2}} = 1 - \frac{1}{2^n} \]

5. **Combine the Results**:
Now substituting back into the equation for \( S_n \):
\[ S_n = n - \left( 1 - \frac{1}{2^n} \right) = n - 1 + \frac{1}{2^n} \]

6. **Final Expression**:
Thus, we can express \( S_n \) as:
\[ S_n = n - 1 + \frac{1}{2^n} \]

7. **Compare with Given Options**:
Now we compare our result with the given options:
- (a) \( 2^n - n - 1 \)
- (b) \( 1 - 2^{-n} \)
- (c) \( 2^{-n} + n - 1 \)
- (d) \( 2^n - 1 \)

Our derived expression \( S_n = n - 1 + \frac{1}{2^n} \) matches option (c) when rearranged:
\[ S_n = 2^{-n} + n - 1 \]

### Conclusion:
The correct answer is **(c) \( 2^{-n} + n - 1 \)**.