The sum ` 1 + 3 + 3^(2) + …+ 3^(n)` is equal to

Options

A. `3^(n-1)`
B. `(3^(n+1) -1)/(2)`
C. `3^(n+1) -1`
D. `3^(n)`

Answer

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

### Step-by-Step Solution:

1. **Identify the first term and common ratio**:
- The first term \( a = 1 \).
- The common ratio \( r = 3 \).

2. **Determine the number of terms**:
- The series goes from \( 3^0 \) to \( 3^n \), which means there are \( n + 1 \) terms in total.

3. **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 progression is:
\[ S_n = a \frac{r^n - 1}{r - 1} \]
- Here, since we have \( n + 1 \) terms, we will use \( n + 1 \) in the formula.

4. **Substitute the values into the formula**:
- Using \( a = 1 \), \( r = 3 \), and \( n + 1 \) terms:
\[ S_{n+1} = 1 \cdot \frac{3^{n+1} - 1}{3 - 1} \]
- This simplifies to:
\[ S_{n+1} = \frac{3^{n+1} - 1}{2} \]

5. **Final Result**:
- Therefore, the sum of the series \( 1 + 3 + 3^2 + \ldots + 3^n \) is:
\[ S = \frac{3^{n+1} - 1}{2} \]