What is the greatest value of the positive integer n satisfying the condition
`1+(1)/(2)+(1)/(4)+(1)/(8)+ . . .+(1)/(2^(n)-1) lt 2-(1)/(1000)0` ?

Options

A. 8
B. 9
C. 10
D. 11

Answer

To solve the problem, we need to find the greatest positive integer \( n \) that satisfies the inequality:

\[ 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots + \frac{1}{2^{n-1}} < 2 - \frac{1}{1000} \]

### Step 1: Identify the left-hand side as a geometric series

The left-hand side is a finite geometric series where:
- The first term \( a = 1 \)
- The common ratio \( r = \frac{1}{2} \)
- The number of terms \( n \)

The sum \( S_n \) of the first \( n \) terms of a geometric series can be calculated using the formula:

\[ S_n = \frac{a(1 - r^n)}{1 - r} \]

### Step 2: Substitute the values into the formula

Substituting the values into the formula, we have:

\[ S_n = \frac{1(1 - (\frac{1}{2})^n)}{1 - \frac{1}{2}} = \frac{1 - \frac{1}{2^n}}{\frac{1}{2}} = 2(1 - \frac{1}{2^n}) = 2 - \frac{2}{2^n} \]

### Step 3: Set up the inequality

Now we set up the inequality:

\[ 2 - \frac{2}{2^n} < 2 - \frac{1}{1000} \]

### Step 4: Simplify the inequality

Subtracting \( 2 \) from both sides gives:

\[ -\frac{2}{2^n} < -\frac{1}{1000} \]

Multiplying through by \( -1 \) (which reverses the inequality):

\[ \frac{2}{2^n} > \frac{1}{1000} \]

### Step 5: Cross-multiply to eliminate the fractions

Cross-multiplying gives:

\[ 2000 > 2^n \]

### Step 6: Solve for \( n \)

To find the maximum \( n \), we need to determine the largest integer \( n \) such that \( 2^n < 2000 \).

Calculating powers of \( 2 \):

- \( 2^{10} = 1024 \)
- \( 2^{11} = 2048 \)

Since \( 1024 < 2000 < 2048 \), the greatest integer \( n \) satisfying this condition is \( n = 10 \).

### Final Answer

Thus, the greatest value of the positive integer \( n \) satisfying the condition is:

\[ \boxed{10} \]