If the sum of `1+2+2^(2)+ . . . . . . . . . .+2^(n-1)` is 255, find the value of n.

Answer

To solve the problem, we need to find the value of \( n \) such that the sum of the series \( 1 + 2 + 2^2 + \ldots + 2^{n-1} \) equals 255.

### Step-by-Step Solution:

1. **Identify the Series**: The series given is a geometric progression (GP) where:
- The first term \( a = 1 \)
- The common ratio \( r = 2 \)

2. **Formula for the Sum of a GP**: The sum \( S_n \) of the first \( n \) terms of a geometric progression can be calculated using the formula:
\[ S_n = \frac{a(r^n - 1)}{r - 1} \]
Here, \( a = 1 \) and \( r = 2 \).

3. **Substituting Values**: Plugging in the values into the formula:
\[ S_n = \frac{1(2^n - 1)}{2 - 1} = 2^n - 1 \]

4. **Setting Up the Equation**: We know from the problem statement that:
\[ S_n = 255 \]
Therefore, we can set up the equation:
\[ 2^n - 1 = 255 \]

5. **Solving for \( 2^n \)**: Adding 1 to both sides gives:
\[ 2^n = 256 \]

6. **Finding \( n \)**: We recognize that \( 256 \) can be expressed as a power of 2:
\[ 256 = 2^8 \]
Thus, we can equate the exponents:
\[ n = 8 \]

### Final Answer:
The value of \( n \) is \( 8 \).

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