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

Options

A. A)`2^(n) -1`
B. B)`1- 2^(-n)`
C. C)`2^n - n +1`
D. D)`n+ 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 follow these steps:

### Step 1: Identify the Pattern in the Series
The series can be rewritten as:
- \( \frac{1}{2} = 1 - \frac{1}{2} \)
- \( \frac{3}{4} = 1 - \frac{1}{4} \)
- \( \frac{7}{8} = 1 - \frac{1}{8} \)
- \( \frac{15}{16} = 1 - \frac{1}{16} \)

From this, we can see that the general term can be expressed as:
\[ \frac{2^n - 1}{2^n} = 1 - \frac{1}{2^n} \]

### Step 2: Write the Sum of the First n Terms
The sum of the first n terms \( S_n \) can be expressed as:
\[ S_n = \left(1 - \frac{1}{2}\right) + \left(1 - \frac{1}{4}\right) + \left(1 - \frac{1}{8}\right) + \ldots + \left(1 - \frac{1}{2^n}\right) \]

This can be simplified as:
\[ S_n = n - \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots + \frac{1}{2^n}\right) \]

### Step 3: Calculate the Sum of the Geometric Series
The series \( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots + \frac{1}{2^n} \) is a geometric series where:
- First term \( a = \frac{1}{2} \)
- Common ratio \( r = \frac{1}{2} \)

The sum of the first n terms of a geometric series is given by:
\[ S = \frac{a(1 - r^n)}{1 - r} \]
Substituting the values:
\[ S = \frac{\frac{1}{2}(1 - (\frac{1}{2})^n)}{1 - \frac{1}{2}} = \frac{\frac{1}{2}(1 - \frac{1}{2^n})}{\frac{1}{2}} = 1 - \frac{1}{2^n} \]

### Step 4: Substitute Back into the Sum
Now substituting back into the expression for \( S_n \):
\[ S_n = n - \left(1 - \frac{1}{2^n}\right) \]
This simplifies to:
\[ S_n = n - 1 + \frac{1}{2^n} \]

### Step 5: Final Expression
Thus, the sum of the first n terms of the series is:
\[ S_n = n + \frac{1}{2^n} - 1 \]

### Conclusion
The final result can be expressed as:
\[ S_n = n + 2^{-n} - 1 \]