The minimum value of n such that `1 + 3 + 3^(2) +...+ 3^(n) gt 1000` is

Options

A. 7
B. 8
C. 9
D. none of these

Answer

To find the minimum value of \( n \) such that the sum \( 1 + 3 + 3^2 + \ldots + 3^n > 1000 \), we can use the formula for the sum of a geometric series.

### Step-by-Step Solution:

1. **Identify the series**: The series is a geometric series where the first term \( a = 1 \) and the common ratio \( r = 3 \). The number of terms in the series is \( n + 1 \).

2. **Write the formula for the sum of the series**: The sum \( S \) of the first \( n + 1 \) terms of a geometric series can be calculated using the formula:
\[ S = \frac{a(r^{n+1} - 1)}{r - 1} \]
Substituting the values of \( a \) and \( r \):
\[ S = \frac{1(3^{n+1} - 1)}{3 - 1} = \frac{3^{n+1} - 1}{2} \]

3. **Set up the inequality**: We need to find \( n \) such that:
\[ \frac{3^{n+1} - 1}{2} > 1000 \]
Multiplying both sides by 2 to eliminate the fraction:
\[ 3^{n+1} - 1 > 2000 \]
Adding 1 to both sides:
\[ 3^{n+1} > 2001 \]

4. **Solve for \( n + 1 \)**: We need to find the smallest integer \( n + 1 \) such that \( 3^{n+1} > 2001 \). We can estimate \( n + 1 \) by calculating powers of 3:
- \( 3^6 = 729 \)
- \( 3^7 = 2187 \)

Since \( 3^6 < 2001 < 3^7 \), it follows that:
\[ n + 1 \geq 7 \]
Therefore, \( n \geq 6 \).

5. **Determine the minimum value of \( n \)**: The minimum integer value of \( n \) that satisfies this condition is:
\[ n = 6 \]

### Conclusion:
Thus, the minimum value of \( n \) such that \( 1 + 3 + 3^2 + \ldots + 3^n > 1000 \) is \( n = 6 \).