Find the sum of given series `1+2+4+8+………+ 256`

Options

A. 510
B. 511
C. 512
D. 513

Explanation

To find the sum of the series \(1 + 2 + 4 + 8 + \ldots + 256\), we can recognize that this is a geometric progression (GP) where each term is obtained by multiplying the previous term by a common ratio.

### Step-by-step solution:

1. **Identify the first term and the common ratio**:
- The first term \(a = 1\).
- The common ratio \(r = 2\) (since each term is double the previous term).

2. **Identify the last term**:
- The last term of the series is \(256\).

3. **Determine the number of terms in the series**:
- The general formula for the \(n\)-th term of a GP is given by:
\[ a_n = a \cdot r^{n-1} \]
- Setting \(a_n = 256\), we have:
\[ 256 = 1 \cdot 2^{n-1} \]
- This simplifies to:
\[ 2^{n-1} = 256 \]
- Since \(256 = 2^8\), we can equate the exponents:
\[ n - 1 = 8 \implies n = 9 \]

4. **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 GP is:
\[ S_n = \frac{a(r^n - 1)}{r - 1} \]
- Substituting the values we have:
\[ S_9 = \frac{1(2^9 - 1)}{2 - 1} \]
- Simplifying further:
\[ S_9 = 2^9 - 1 = 512 - 1 = 511 \]

5. **Final answer**:
- The sum of the series \(1 + 2 + 4 + 8 + \ldots + 256\) is \(511\).