If harmonic mean of `(1)/(2),(1)/(2^(2)),(1)/(2^(3)),...,(1)/(2^(10))` is `(lambda)/(2^(10)-1)`, then `lambda=`

Options

A. `10.2^(10)`
B. `5`
C. `5.2^(10)`
D. `10`

Explanation

To solve the problem, we need to find the harmonic mean of the series \( \frac{1}{2}, \frac{1}{2^2}, \frac{1}{2^3}, \ldots, \frac{1}{2^{10}} \) and express it in the form \( \frac{\lambda}{2^{10}-1} \).

### Step-by-step Solution:

1. **Identify the Series**:
The series given is:
\[ a_1 = \frac{1}{2}, a_2 = \frac{1}{2^2}, a_3 = \frac{1}{2^3}, \ldots, a_{10} = \frac{1}{2^{10}} \]

2. **Formula for Harmonic Mean**:
The harmonic mean \( H \) of \( n \) numbers \( a_1, a_2, \ldots, a_n \) is given by:
\[ H = \frac{n}{\frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3} + \ldots + \frac{1}{a_n}} \]
In our case, \( n = 10 \).

3. **Calculate the Denominator**:
We need to calculate:
\[ \frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3} + \ldots + \frac{1}{a_{10}} = 2 + 2^2 + 2^3 + \ldots + 2^{10} \]
This is a geometric series where the first term \( a = 2 \) and the common ratio \( r = 2 \).

4. **Sum of the Geometric Series**:
The sum \( S \) of the first \( n \) terms of a geometric series is given by:
\[ S = a \frac{r^n - 1}{r - 1} \]
Here, \( a = 2 \), \( r = 2 \), and \( n = 10 \):
\[ S = 2 \frac{2^{10} - 1}{2 - 1} = 2(2^{10} - 1) = 2^{11} - 2 \]

5. **Substituting Back into the Harmonic Mean Formula**:
Now substituting back into the harmonic mean formula:
\[ H = \frac{10}{2^{11} - 2} \]

6. **Simplifying the Harmonic Mean**:
We can factor out 2 in the denominator:
\[ H = \frac{10}{2(2^{10} - 1)} = \frac{5}{2^{10} - 1} \]

7. **Comparing with the Given Form**:
We are given that the harmonic mean is \( \frac{\lambda}{2^{10} - 1} \). From our calculation, we have:
\[ H = \frac{5}{2^{10} - 1} \]
Thus, we can see that \( \lambda = 5 \).

### Final Answer:
\[ \lambda = 5 \]