The least value of n, such that (`1+3+3^2+……….+3^n)` exceeds 2000 is

Options

A. 5
B. 6
C. 7
D. 8

Answer

To find the least value of \( n \) such that the sum \( S = 1 + 3 + 3^2 + \ldots + 3^n \) exceeds 2000, we can recognize that this is a geometric series.

### Step-by-Step Solution:

1. **Identify the formula for the sum of a geometric series**:
The sum \( S \) of a geometric series can be calculated using the formula:
\[ S_n = \frac{a(r^{n+1} - 1)}{r - 1} \]
where \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms. In our case, \( a = 1 \), \( r = 3 \), and there are \( n + 1 \) terms.

2. **Apply the formula**:
Plugging in the values, we get:
\[ S_n = \frac{1(3^{n+1} - 1)}{3 - 1} = \frac{3^{n+1} - 1}{2} \]

3. **Set up the inequality**:
We need \( S_n \) to exceed 2000:
\[ \frac{3^{n+1} - 1}{2} > 2000 \]

4. **Multiply both sides by 2**:
\[ 3^{n+1} - 1 > 4000 \]

5. **Add 1 to both sides**:
\[ 3^{n+1} > 4001 \]

6. **Determine the smallest \( n+1 \)**:
We need to find the smallest integer \( n+1 \) such that \( 3^{n+1} > 4001 \). We can calculate powers of 3:
- \( 3^6 = 729 \)
- \( 3^7 = 2187 \)
- \( 3^8 = 6561 \)

Since \( 3^8 = 6561 \) is the first power greater than 4001, we have:
\[ n + 1 = 8 \]

7. **Solve for \( n \)**:
\[ n = 8 - 1 = 7 \]

### Conclusion:
The least value of \( n \) such that \( 1 + 3 + 3^2 + \ldots + 3^n \) exceeds 2000 is \( n = 7 \).