The sum of n terms the series `1+1/2+1/2^2+1/2^3+……….`is

Options

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

Answer

To find the sum of the first n terms of the series \(1 + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \ldots\), we can recognize that this is a geometric progression (GP).

### Step-by-Step Solution:

1. **Identify the first term (a) and the common ratio (r)**:
- The first term \(a = 1\).
- The common ratio \(r = \frac{1}{2}\).

2. **Use the formula for the sum of the first n terms of a GP**:
The formula for the sum \(S_n\) of the first n terms of a geometric series is:
\[ S_n = \frac{a(1 - r^n)}{1 - r} \]

3. **Substitute the values of a and r into the formula**:
\[ S_n = \frac{1(1 - (\frac{1}{2})^n)}{1 - \frac{1}{2}} \]

4. **Simplify the denominator**:
The denominator \(1 - \frac{1}{2} = \frac{1}{2}\).

5. **Rewrite the equation**:
\[ S_n = \frac{1(1 - (\frac{1}{2})^n)}{\frac{1}{2}} = 2(1 - (\frac{1}{2})^n) \]

6. **Distribute the 2**:
\[ S_n = 2 - 2(\frac{1}{2})^n = 2 - \frac{2}{2^n} \]

7. **Combine the terms**:
\[ S_n = 2 - \frac{2}{2^n} = \frac{2 \cdot 2^n - 2}{2^n} = \frac{2^{n+1} - 2}{2^n} \]

8. **Final expression**:
\[ S_n = \frac{2(2^n - 1)}{2^n} \]

### Final Answer:
The sum of the first n terms of the series is:
\[ S_n = 2(1 - \frac{1}{2^n}) = 2 - \frac{2}{2^n} \]