Find the least value of n for which `1+3+3^(2)+…3^(n-1)>1000`

Answer

To find the least value of \( n \) for which the sum \( 1 + 3 + 3^2 + \ldots + 3^{n-1} > 1000 \), we can use the formula for the sum of a geometric series.

### Step-by-step Solution:

1. **Identify the Series**:
The series can be rewritten as:
\[ 1 + 3 + 3^2 + \ldots + 3^{n-1} \]
This is a geometric series where the first term \( a = 1 \) and the common ratio \( r = 3 \).

2. **Use the Sum Formula for a Geometric Series**:
The sum \( S_n \) of the first \( n \) terms of a geometric series is given by:
\[ S_n = \frac{a(r^n - 1)}{r - 1} \]
Substituting the values of \( a \) and \( r \):
\[ S_n = \frac{1(3^n - 1)}{3 - 1} = \frac{3^n - 1}{2} \]

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

4. **Estimate \( n \)**:
We can find \( n \) by calculating powers of 3:
- \( 3^6 = 729 \) (which is less than 2001)
- \( 3^7 = 2187 \) (which is greater than 2001)

Since \( 3^6 < 2001 < 3^7 \), the least integer \( n \) that satisfies the inequality is:
\[ n = 7 \]

### Final Answer:
The least value of \( n \) for which \( 1 + 3 + 3^2 + \ldots + 3^{n-1} > 1000 \) is \( n = 7 \).